The post इंडिया गेट पर निबंध – Essay on India Gate in Hindi appeared first on Babaji Academy.

]]>विश्व युद्ध के दौरान मारे गए भारत के सैनिकों को सम्मानित करने के लिए इंडिया गेट का निर्माण किया गया था। यह संरचना भारत की एक सुंदर वास्तुकला का दावा करती है। इसका उद्घाटन पहली बार वर्ष 1931 में किया गया था और इसमें अभी भी वही आकर्षण है। साथ ही, यह संरचना हमें उन हजारों सैनिकों द्वारा किए गए बलिदानों की याद दिलाती है, जिन्होंने प्रथम विश्व युद्ध के दौरान अपना जीवन दिया है। इंडिया गेट पर निबंध भारत के इस प्रसिद्ध स्मारक की अंतर्दृष्टि है।

जब भी कोई दिल्ली जाता है, तो वे कभी भी इंडिया गेट को मिस नहीं करते हैं। इस प्रकार, यह दिल्ली के अभिन्न हिस्सों में से एक बन गया है और एक प्रमुख पर्यटक आकर्षण भी है। स्थानीय लोगों के साथ-साथ दिल्ली घूमने आने वाले पर्यटक साल भर इस स्थान पर आते हैं। इसके अलावा, कई स्थानीय लोगों के लिए, दोनों किनारों पर गेट के चारों ओर फैले लॉन एक अच्छा पिकनिक स्थल है।

भारत के द्वार तक पहुंचना बहुत आसान है क्योंकि यह केंद्र में स्थित है। इसके अलावा, सर्दियों के लिए, लोग अपने दोपहर और रात के खाने के लिए अपने दोस्तों और परिवारों के साथ इंडिया गेट पर जाने के लिए क्वालिटी टाइम बिताते हैं। इन बड़ों के साथ बैठे बड़ों के साथ बच्चे खेल सकते हैं। गर्मी के महीनों के दौरान यह नजारा भी अच्छा लगता है। हालांकि, इस जगह की यात्रा के लिए सबसे खूबसूरत समय रात में होता है जब यह रोशनी से पूरी तरह से रोशन होता है।

इंडिया गेट एक स्मारक है जिसे प्रथम विश्व युद्ध (1914- 1918) में शहीद हुए सैनिकों को याद करने के लिए बनाया गया था। यह नई दिल्ली में राजपथ पर स्थित है और सर एडविन लुटियन द्वारा डिजाइन किया गया था। यह प्रसिद्ध है क्योंकि यह भारत का सबसे बड़ा युद्ध स्मारक है। यह कई कारणों से भारत में सबसे अधिक देखी जाने वाली जगहों में से एक है।

भारत हर जगह एक स्मारक के साथ-साथ एक अद्भुत वास्तुशिल्प डिजाइन के रूप में प्रसिद्ध है। इस स्मारक को बनाने में 10 साल लगे और इसे एडविन लुटियंस की देखरेख में तैयार किया गया था। इसके अलावा, वह शाही युद्ध कब्र आयोग का सदस्य था और दिसंबर 1917 में इसका गठन किया गया था। इसके अलावा, वह युद्ध स्मारक और कब्रों के डिजाइन में विशेष था। इस प्रकार, एडविन लुटियंस को भारत में युद्ध स्मारक को डिजाइन करने का यह कार्य दिया गया था।

इंडिया गेट दिल्ली के बीचोबीच स्थित है और 42 मीटर लंबी एक इमारत है जो चौड़ाई 9.1 मीटर है। इसके अलावा, गेट मुख्य रूप से लाल और पीले बलुआ पत्थर और ग्रेनाइट से बना है। इसके अतिरिक्त, भारत गेट की स्थापत्य शैली विजयी मेहराब पर आधारित है। इसलिए, इंडिया गेट के शीर्ष पर गुंबददार कटोरी है और इसे वर्षगांठ और राष्ट्रीय त्योहारों जैसे महत्वपूर्ण दिनों में जलते हुए तेल के साथ फिल्माने के उद्देश्य से बनाया गया है । साथ ही, इंडिया गेट का वास्तुशिल्प आर्क डी ट्रायम्फे डी ल ईटोली पर आधारित है। यह पेरिस के सबसे प्रसिद्ध स्मारकों में से एक माना जाता है।

प्रथम भारतीय युद्ध के दौरान मारे गए ब्रिटिश भारतीय के सैनिकों को याद करने के लिए इंडिया गेट बनाया गया था, एक छोटी सी इमारत थी जो 1971 में बांग्लादेश मुक्ति युद्ध के दौरान अपनी जान गंवाने वाले सैनिकों के सम्मान में बनाई गई थी। इस प्रकार, अमर जवान ज्योति को भारत द्वार का एक अभिन्न अंग माना जाता है। अमर जवान ज्योति को शीर्ष पर काले हेलमेट के साथ कवर की गई राइफल के साथ काले पैडल के रूप में देखा जाता है।

The post इंडिया गेट पर निबंध – Essay on India Gate in Hindi appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 7 – Alternating Current appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated Class 12 Physics solutions of Chapter 6 i.e Electromagnetic Induction. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current**

Section Name | Topic Name |

7 | Alternating Current |

7.1 | Introduction |

7.2 | AC Voltage Applied to a Resistor |

7.3 | Representation of AC Current and Voltage by Rotating Vectors — Phasors |

7.4 | AC Voltage Applied to an Inductor |

7.5 | AC Voltage Applied to a Capacitor |

7.6 | AC Voltage Applied to a Series LCR Circuit |

7.7 | Power in AC Circuit: The Power Factor |

7.8 | LC Oscillations |

7.9 | Transformers |

A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.

**(a)** What is the rms value of current in the circuit?

**(b) **What is the net power consumed over a full cycle?

Resistance of the resistor, *R *= 100 Ω

Supply voltage, *V* = 220 V

Frequency, *ν*** = **50 Hz

**(a)** The rms value of current in the circuit is given as:

**(b) **The net power consumed over a full cycle is given as:

*P* = *VI*

= 220 × 2.2 = 484 W

**(a)** The peak voltage of an ac supply is 300 V. What is the rms voltage?

**(b)** The rms value of current in an ac circuit is 10 A. What is the peak current?

**(a)** Peak voltage of the ac supply, *V*_{0} = 300 V

Rms voltage is given as:

**(b)** Therms value of current is given as:

*I* = 10 A

Now, peak current is given as:

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Inductance of inductor, *L *= 44 mH = 44 × 10^{−3} H

Supply voltage, *V* = 220 V

Frequency, *ν* = 50 Hz

Angular frequency, *ω*=

Inductive reactance, *X*_{L} = *ω* *L*

Rms value of current is given as:

Hence, the rms value of current in the circuit is 15.92 A.

A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Capacitance of capacitor, *C* = 60 μF = 60 × 10^{−6} F

Supply voltage, *V *= 110 V

Frequency, *ν* = 60 Hz

Angular frequency, *ω*=

Capacitive reactance

Rms value of current is given as:

Hence, the rms value of current is 2.49 A.

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

In the inductive circuit,

Rms value of current,* I* = 15.92 A

Rms value of voltage, *V* = 220 V

Hence, the net power absorbed can be obtained by the relation,

*P* = *VI* cos *Φ*

Where,

Φ = Phase difference between *V* and *I*

For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., *Φ*= 90°.

Hence, *P* = 0 i.e., the net power is zero.

In the capacitive circuit,

Rms value of current, *I* = 2.49 A

Rms value of voltage, *V* = 110 V

Hence, the net power absorbed can ve obtained as:

*P* = *VI *Cos *Φ*

For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., *Φ*= 90°.

Hence,* P *= 0 i.e., the net power is zero.

Obtain the resonant frequency *ω**r *of a series *LCR *circuit with *L *= 2.0 H, *C *= 32 μF and *R *= 10 Ω. What is the *Q*-value of this circuit?

Inductance, *L *= 2.0 H

Capacitance, *C* = 32 μF = 32 × 10^{−6} F

Resistance, *R* = 10 Ω

Resonant frequency is given by the relation,

Now, *Q*-value of the circuit is given as:

Hence, the Q-Value of this circuit is 25.

A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Capacitance, *C* = 30μF = 30×10^{−6}F

Inductance, *L* = 27 mH = 27 × 10^{−3} H

Angular frequency is given as:

Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10^{3} rad/s.

Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Capacitance of the capacitor, *C* = 30 μF = 30×10^{−6 }F

Inductance of the inductor, *L* = 27 mH = 27 × 10^{−3} H

Charge on the capacitor, *Q* = 6 mC = 6 × 10^{−3} C

Total energy stored in the capacitor can be calculated by the relation,

Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.

A series *LCR *circuit with *R *= 20 Ω, *L *= 1.5 H and *C *= 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

At resonance, the frequency of the supply power equals the natural frequency of the given *LCR* circuit.

Resistance, *R* = 20 Ω

Inductance, *L* = 1.5 H

Capacitance, *C* = 35 μF = 30 × 10^{−6 }F

AC supply voltage to the *LCR* circuit, *V* = 200 V

Impedance of the circuit is given by the relation,

At resonance,

Current in the circuit can be calculated as:

Hence, the average power transferred to the circuit in one complete cycle= *VI*

= 200 × 10 = 2000 W.

A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its *LC *circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?[*Hint: *For tuning, the natural frequency i.e., the frequency of free oscillations of the *LC *circuit should be equal to the frequency of the radiowave.]

The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.

Lower tuning frequency, ν_{1} = 800 kHz = 800 × 10^{3} Hz

Upper tuning frequency, ν_{2} = 1200 kHz = 1200× 10^{3} Hz

Effective inductance of circuit *L* = 200 μH = 200 × 10^{−6} H

Capacitance of variable capacitor for ν_{1 }is given as_{:}

*C*_{1}

Where,

ω_{1} = Angular frequency for capacitor *C*_{1}

Capacitance of variable capacitor for ν_{2,}

*C*_{2}

Where,

ω_{2} = Angular frequency for capacitor *C*_{2}

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

Figure 7.21 shows a series *LCR *circuit connected to a variable frequency 230 V source. *L *= 5.0 H, *C *= 80μF, *R *= 40 Ω

**(a) **Determine the source frequency which drives the circuit in resonance.

**(b) **Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

**(c) **Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the *LC *combination is zero at the resonating frequency.

Inductance of the inductor, *L *= 5.0 H

Capacitance of the capacitor, *C* = 80 μF = 80 × 10^{−6} F

Resistance of the resistor, *R* = 40 Ω

Potential of the variable voltage source, *V* = 230 V

**(a)** Resonance angular frequency is given as:

Hence, the circuit will come in resonance for a source frequency of 50 rad/s.

**(b)** Impedance of the circuit is given by the relation,

At resonance,

Amplitude of the current at the resonating frequency is given as:

Where,

*V*_{0} = Peak voltage

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.

**(c)** Rms potential drop across the inductor,

(*V*_{L})_{rms} = *I* × *ω*_{R}*L*

Where,

*I* = rms current

Potential drop across the capacitor,

Potential drop across the resistor,

(*V*_{R})_{rms} = *IR*

= × 40 = 230 V

Potential drop across the *LC *combination,

At resonance,

∴*V*_{LC}= 0

Hence, it is proved that the potential drop across the *LC *combination is zero at resonating frequency.

An *LC *circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be *t *= 0.

**(a) **What is the total energy stored initially? Is it conserved during *LC *oscillations?

**(b) **What is the natural frequency of the circuit?

**(c) **At what time is the energy stored

(i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)?

**(d) **At what times is the total energy shared equally between the inductor and the capacitor?

**(e) **If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Inductance of the inductor, *L* = 20 mH = 20 × 10^{−3} H

Capacitance of the capacitor, *C* = 50 μF = 50 × 10^{−6} F

Initial charge on the capacitor, *Q* = 10 mC = 10 × 10^{−3} C

**(a)** Total energy stored initially in the circuit is given as:

Hence, the total energy stored in the *LC *circuit will be conserved because there is no resistor connected in the circuit.

**(b)**Natural frequency of the circuit is given by the relation,

Natural angular frequency,

Hence, the natural frequency of the circuit is 10^{3} rad/s.

**(c) (i) **For time period (*T*), total charge on the capacitor at time *t*,

For energy stored is electrical, we can write *Q*’ = *Q*.

Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, *t* =

**(ii) **Magnetic energy is the maximum when electrical energy, *Q*′ is equal to 0.

Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time,

**(d)** *Q*^{1} = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time *t*.

When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = (maximum energy).

Hence, total energy is equally shared between the inductor and the capacity at time,

**(e)** If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the *LC* oscillation.

A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.

**(a) **What is the maximum current in the coil?

**(b) **What is the time lag between the voltage maximum and the current maximum?

Inductance of the inductor,* L *= 0.50 H

Resistance of the resistor, *R* = 100 Ω

Potential of the supply voltage, *V* = 240 V

Frequency of the supply, *ν* = 50 Hz

**(a) **Peak voltage is given as:

Angular frequency of the supply,

ω = 2 π*ν*

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

**(b)** Equation for voltage is given as:

*V* = *V*_{0} cos *ω**t*

Equation for current is given as:

*I* = *I*_{0} cos (*ω**t* − *Φ*)

Where,

Φ = Phase difference between voltage and current

At time, *t *= 0.

*V* = *V*_{0}(voltage is maximum)

For*ω**t* − *Φ* = 0 i.e., at time,

*I* = *I*_{0} (current is maximum)

Hence, the time lag between maximum voltage and maximum current is.

Now, phase angle *Φ*is given by the relation,

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Inductance of the inductor, *L* = 0.5 Hz

Resistance of the resistor, *R* = 100 Ω

Potential of the supply voltages, *V* = 240 V

Frequency of the supply,*ν*** = **10 kHz = 10^{4} Hz

Angular frequency, *ω* = 2π*ν*= 2π × 10^{4} rad/s

**(a) **Peak voltage,

Maximum current,

**(b) **For phase difference*Φ*, we have the relation:

It can be observed that *I*_{0} is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, *ω* = 0. Hence, inductor L behaves like a pure conducting object.

A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

**(a) **What is the maximum current in the circuit?

**(b) **What is the time lag between the current maximum and the voltage maximum?

Capacitance of the capacitor, *C* = 100 μF = 100 × 10^{−6} F

Resistance of the resistor, *R* = 40 Ω

Supply voltage, *V* = 110 V

**(a)** Frequency of oscillations, *ν*= 60 Hz

Angular frequency,

For a *RC* circuit, we have the relation for impedance as:

Peak voltage, *V*_{0} =

Maximum current is given as:

**(b) **In a capacitor circuit, the voltage lags behind the current by a phase angle of*Φ*. This angle is given by the relation:

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.

Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Capacitance of the capacitor, *C* = 100 μF = 100 × 10^{−6} F

Resistance of the resistor,* R* = 40 Ω

Supply voltage, *V* = 110 V

Frequency of the supply, *ν* = 12 kHz = 12 × 10^{3} Hz

Angular Frequency, *ω* = 2 π*ν*= 2 × π × 12 × 10^{3}03

= 24π × 10^{3} rad/s

Peak voltage,

Maximum current,

For an *RC* circuit, the voltage lags behind the current by a phase angle of *Φ* given as:

Hence, *Φ* tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.

In a dc circuit, after the steady state is achieved, *ω* = 0. Hence, capacitor C amounts to an open circuit.

Keeping the source frequency equal to the resonating frequency of the series *LCR *circuit, if the three elements, *L*, *C *and *R *are arranged in parallel, show that the total current in the parallel *LCR *circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

An inductor (*L*), a capacitor (*C*), and a resistor (*R*) is connected in parallel with each other in a circuit where,

*L* = 5.0 H

*C* = 80 μF = 80 × 10^{−6 }F

*R* = 40 Ω

Potential of the voltage source, *V* = 230 V

Impedance (Z) of the given parallel *LCR* circuit is given as:

Where,

ω = Angular frequency

At resonance,

Hence, the magnitude of *Z *is the maximum at 50 rad/s. As a result, the total current is minimum.

Rms current flowing through inductor L is given as:

Rms current flowing through capacitor C is given as:

Rms current flowing through resistor R is given as:

A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.

**(a) **Obtain the current amplitude and rms values.

**(b) **Obtain the rms values of potential drops across each element.

**(c) **What is the average power transferred to the inductor?

**(d) **What is the average power transferred to the capacitor?

**(e) **What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Inductance, *L* = 80 mH = 80 × 10^{−3} H

Capacitance, *C *= 60 μF = 60 × 10^{−6 }F

Supply voltage, *V* = 230 V

Frequency, *ν* = 50 Hz

Angular frequency, *ω* = 2π*ν*= 100 π rad/s

Peak voltage, *V*_{0}=

**(a)** Maximum current is given as:

The negative sign appears because

Amplitude of maximum current,

Hence, rms value of current,

**(b)** Potential difference across the inductor,

*V*_{L}=* I *× ω*L*

= 8.22 × 100 π × 80 × 10^{−3}

= 206.61 V

Potential difference across the capacitor,

**(c)** Average power consumed by the inductor is zero as actual voltage leads the current by.

**(d)** Average power consumed by the capacitor is zero as voltage lags current by.

**(e)** The total power absorbed (averaged over one cycle) is zero.

Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Average power transferred to the resistor = 788.44 W

Average power transferred to the capacitor = 0 W

Total power absorbed by the circuit = 788.44 W

Inductance of inductor, *L* = 80 mH = 80 × 10^{−3} H

Capacitance of capacitor, *C* = 60 *μ*F = 60 × 10^{−6} F

Resistance of resistor, *R* = 15 Ω

Potential of voltage supply, *V* = 230 V

Frequency of signal, *ν* = 50 Hz

Angular frequency of signal, *ω* = 2π*ν*= 2π × (50) = 100π rad/s

The elements are connected in series to each other. Hence, impedance of the circuit is given as:

Current flowing in the circuit,

Average power transferred to resistance is given as:

*P*_{R}= *I*^{2}*R*

= (7.25)^{2} × 15 = 788.44 W

Average power transferred to capacitor, *P*_{C} = Average power transferred to inductor, *P*_{L} = 0

Total power absorbed by the circuit:

= *P*_{R }*+ P*_{C}* + P*_{L}

= 788.44 + 0 + 0 = 788.44 W

Hence, the total power absorbed by the circuit is 788.44 W.

A series *LCR *circuit with *L *= 0.12 H, *C *= 480 nF, *R *= 23 Ω is connected to a 230 V variable frequency supply.

**(a) **What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

**(b) **What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

**(c)** For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

**(d) **What is the *Q*-factor of the given circuit?

Inductance, *L* = 0.12 H

Capacitance, *C* = 480 nF = 480 × 10^{−9} F

Resistance, *R* = 23 Ω

Supply voltage, *V* = 230 V

Peak voltage is given as:

*V*_{0} = = 325.22 V

**(a)** Current flowing in the circuit is given by the relation,

Where,

*I*_{0} = maximum at resonance

At resonance, we have

Where,

ω_{R }*= *Resonance angular frequency

∴Resonant frequency,

And, maximum current

**(b)** Maximum average power absorbed by the circuit is given as:

Hence, resonant frequency () is

**(c)** The power transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half, =

Where,

Hence, change in frequency,

∴

And,

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

**(d)** *Q*-factor of the given circuit can be obtained using the relation,

Hence, the Q-factor of the given circuit is 21.74.

Obtain the resonant frequency and *Q*-factor of a series *LCR *circuit with *L *= 3.0 H, *C *= 27 μF, and *R *= 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.

Inductance, *L* = 3.0 H

Capacitance, *C* = 27 μF = 27 × 10^{−6} F

Resistance, *R* = 7.4 Ω

At resonance, angular frequency of the source for the given *LCR* series circuit is given as:

*Q*-factor of the series:

To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing, we need to reduce *R* to half i.e.,

Resistance =

Answer the following questions:

**(a) **In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

**(b) **A capacitor is used in the primary circuit of an induction coil.

**(c) **An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across *C *and the ac signal across *L*.

**(d) **A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

**(e) **Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

**(a)** Yes; the statement is not true for rms voltage

It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements may not be in phase.

**(b)** High induced voltage is used to charge the capacitor.

A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.

**(c)** The dc signal will appear across capacitor *C* because for dc signals, the impedance of an inductor (*L*) is negligible while the impedance of a capacitor (*C*) is very high (almost infinite). Hence, a dc signal appears across *C*. For an ac signal of high frequency, the impedance of *L* is high and that of *C* is very low. Hence, an ac signal of high frequency appears across *L*.

**(d)** If an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core increase the impedance of the circuit.

**(e)** A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of a choke coil for this purpose because it wastes power in the form of heat.

A power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?

Input voltage, *V*_{1} = 2300

Number of turns in primary coil, *n*_{1} = 4000

Output voltage, *V*_{2} = 230 V

Number of turns in secondary coil = *n*_{2}

Voltage is related to the number of turns as:

Hence, there are 400 turns in the second winding.

At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m^{3 }s^{−1}. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g= 9.8 m s^{−2}).

Height of water pressure head, *h* = 300 m

Volume of water flow per second, *V* = 100 m^{3}/s

Efficiency of turbine generator, *n* = 60% = 0.6

Acceleration due to gravity, g = 9.8 m/s^{2}

Density of water, *ρ* = 10^{3} kg/m^{3}

Electric power available from the plant = *η* × *hρ*g*V*

= 0.6 × 300 × 10^{3} × 9.8 × 100

= 176.4 × 10^{6} W

= 176.4 MW

A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.

**(a)** Estimate the line power loss in the form of heat.

**(b)** How much power must the plant supply, assuming there is negligible power loss due to leakage?

**(c)** Characterise the step up transformer at the plant.

Total electric power required, *P* = 800 kW = 800 × 10^{3} W

Supply voltage, *V* = 220 V

Voltage at which electric plant is generating power, *V*‘ = 440 V

Distance between the town and power generating station, *d* = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wires, *R* = (15 + 15)0.5 = 15 Ω

A step-down transformer of rating 4000 − 220 V is used in the sub-station.

Input voltage, *V*_{1} = 4000 V

Output voltage, *V*_{2} = 220 V

Rms current in the wire lines is given as:

**(a)** Line power loss = *I*^{2}*R*

= (200)^{2} × 15

= 600 × 10^{3} W

= 600 kW

**(b)** Assuming that the power loss is negligible due to the leakage of the current:

Total power supplied by the plant = 800 kW + 600 kW

= 1400 kW

**(c)** Voltage drop in the power line = *IR* = 200 × 15 = 3000 V

Hence, total voltage transmitted from the plant = 3000 + 4000

= 7000 V

Also, the power generated is 440 V.

Hence, the rating of the step-up transformer situated at the power plant is 440 V − 7000 V.

Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

The rating of a step-down transformer is 40000 V−220 V.

Input voltage, *V*_{1} = 40000 V

Output voltage, *V*_{2} = 220 V

Total electric power required, *P* = 800 kW = 800 × 10^{3} W

Source potential, *V* = 220 V

Voltage at which the electric plant generates power, *V*‘ = 440 V

Distance between the town and power generating station, *d* = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wire lines, *R* = (15 + 15)0.5 = 15 Ω

*P* = V_{1}*I*

Rms current in the wire line is given as:

**(a)** Line power loss = *I*^{2}*R*

= (20)^{2} × 15

= 6 kW

**(b)** Assuming that the power loss is negligible due to the leakage of current.

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW

**(c)** Voltage drop in the power line = *IR *= 20 × 15 = 300 V

Hence, voltage that is transmitted by the power plant

= 300 + 40000 = 40300 V

The power is being generated in the plant at 440 V.

Hence, the rating of the step-up transformer needed at the plant is

440 V − 40300 V.

Hence, power loss during transmission =

In the previous exercise, the power loss due to the same reason is. Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.

**Conclusion**: If you liked** NCERT Class 12 Physics Solutions Chapter 7 – Alternating Current**, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 7 – Alternating Current appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 6 – Electromagnetic Induction appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated Class 12 Physics solutions of Chapter 5 i.e Magnetism and Matter. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction**

Section Name | Topic Name |

6 | Electromagnetic Induction |

6.1 | Introduction |

6.2 | The Experiments of Faraday and Henry |

6.3 | Magnetic Flux |

6.4 | Faraday’s Law of Induction |

6.5 | Lenz’s Law and Conservation of Energy |

6.6 | Motional Electromotive Force |

6.7 | Energy Consideration: A Quantitative Study |

6.8 | Eddy Currents |

6.9 | Inductance |

6.10 | AC Generator |

Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).

**(a)**

**(b)**

**(c)**

**(d)**

**(e)**

**(f)**

The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:

**(a)** The direction of the induced current is along **qrpq***.*

**(b)** The direction of the induced current is along **prqp**.

**(c)** The direction of the induced current is along * yzxy*.

**(d)** The direction of the induced current is along * zyxz*.

**(e)** The direction of the induced current is along * xryx*.

**(f) **No current is induced since the field lines are lying in the plane of the closed loop.

Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:

**(a) **A wire of irregular shape turning into a circular shape;

**(b) **A circular loop being deformed into a narrow straight wire.

According to Lenz’s law, the direction of the induced *emf* is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

**(a) **When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.

**(b) **When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm^{2 }placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m

Number of turns per unit length, *n *= 1500 turns

The solenoid has a small loop of area, *A* = 2.0 cm^{2} = 2 × 10^{−4} m^{2}

Current carried by the solenoid changes from 2 A to 4 A.

Change in current in the solenoid, *di *= 4 − 2 = 2 A

Change in time, *dt* = 0.1 s

Induced *emf* in the solenoid is given by Faraday’s law as:

Where,

= Induced flux through the small loop

= *BA *… (*ii*)

*B *= Magnetic field

=

μ_{0} = Permeability of free space

*= *4π×10^{−7} H/m

Hence, equation (*i*) reduces to:

Hence, the induced voltage in the loop is

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^{−1} in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Length of the rectangular wire, *l* = 8 cm = 0.08 m

Width of the rectangular wire, *b* = 2 cm = 0.02 m

Hence, area of the rectangular loop,

*A* =* lb*

= 0.08 × 0.02

= 16 × 10^{−4} m^{2}

Magnetic field strength, *B* = 0.3 T

Velocity of the loop, *v* = 1 cm/s = 0.01 m/s

**(a) **Emf developed in the loop is given as:

*e* = *Blv*

= 0.3 × 0.08 × 0.01 = 2.4 × 10^{−4} V

Hence, the induced voltage is 2.4 × 10^{−4} V which lasts for 2 s.

**(b) **Emf developed,* e* = *Bbv*

= 0.3 × 0.02 × 0.01 = 0.6 × 10^{−4} V

Hence, the induced voltage is 0.6 × 10^{−4} V which lasts for 8 s.

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^{−1 }about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Length of the rod, *l* = 1 m

Angular frequency,*ω* = 400 rad/s

Magnetic field strength, *B* = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of *l**ω**.*

Average linear velocity of the rod,

Emf developed between the centre and the ring,

Hence, the emf developed between the centre and the ring is 100 V.

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^{−1} in a uniform horizontal magnetic field of magnitude 3.0×10^{−2} T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Max induced *emf *= 0.603 V

Average induced *emf* = 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power comes from the external rotor)

Radius of the circular coil, *r* = 8 cm = 0.08 m

Area of the coil, *A* = π*r*^{2} = π × (0.08)^{2 }m^{2}

Number of turns on the coil, *N* = 20

Angular speed, *ω* = 50 rad/s

Magnetic field strength, *B* = 3 × 10^{−2} T

Resistance of the loop, *R* = 10 Ω

Maximum induced *emf* is given as:

*e* = *N**ω** AB*

= 20 × 50 × π × (0.08)^{2} × 3 × 10^{−2}

= 0.603 V

The maximum *emf* induced in the coil is 0.603 V.

Over a full cycle, the average *emf* induced in the coil is zero.

Maximum current is given as:

Average power loss due to joule heating:

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s^{−1}, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^{−4} Wb m^{−2}.

**(a) **What is the instantaneous value of the emf induced in the wire?

**(b) **What is the direction of the emf?

**(c) **Which end of the wire is at the higher electrical potential?

Length of the wire, *l* = 10 m

Falling speed of the wire, *v* = 5.0 m/s

Magnetic field strength, *B* = 0.3 × 10^{−4} Wb m^{−2}

**(a) **Emf induced in the wire,

*e* = *Blv*

**(b) **Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from West to East.

**(c) **The eastern end of the wire is at a higher potential.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Initial current, *I*_{1} = 5.0 A

Final current, *I*_{2} = 0.0 A

Change in current,

Time taken for the change, *t* = 0.1 s

Average emf, *e* = 200 V

For self-inductance (*L)* of the coil, we have the relation for average emf as:

*e* = *L*

Hence, the self induction of the coil is 4 H.

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Mutual inductance of a pair of coils, *µ* = 1.5 H

Initial current, *I*_{1} = 0 A

Final current *I*_{2} = 20 A

Change in current,

Time taken for the change, *t* = 0.5 s

Induced emf,

Where is the change in the flux linkage with the coil.

Emf is related with mutual inductance as:

Equating equations (1) and (2), we get

Hence, the change in the flux linkage is 30 Wb.

A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^{−4} T and the dip angle is 30°.

Speed of the jet plane, *v* = 1800 km/h = 500 m/s

Wing spanof jet plane,* l *= 25 m

Earth’s magnetic field strength, *B* = 5.0 × 10^{−4} T

Angle of dip,

Vertical component of Earth’s magnetic field,

*B*_{V} = *B* sin

= 5 × 10^{−4} sin 30°

= 2.5 × 10^{−4} T

Voltage difference between the ends of the wing can be calculated as:

*e* = (*B*_{V}) × *l × v*

= 2.5 × 10^{−4} × 25 × 500

= 3.125 V

Hence, the voltage difference developed between the ends of the wings is

3.125 V.

Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^{−1}. If the cut is joined and the loop has a resistance of 1.6 Ω how much power is dissipated by the loop as heat? What is the source of this power?

Sides of the rectangular loop are 8 cm and 2 cm.

Hence, area of the rectangular wire loop,

*A* = length × width

= 8 × 2 = 16 cm^{2}

= 16 × 10^{−4} m^{2}

Initial value of the magnetic field,

Rate of decrease of the magnetic field,

*Emf* developed in the loop is given as:

Where,

= Change in flux through the loop area

= *AB*

Resistance of the loop, *R* = 1.6 Ω

The current induced in the loop is given as:

Power dissipated in the loop in the form of heat is given as:

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^{−1} in the positive *x-*direction in an environment containing a magnetic field in the positive *z*-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^{−3} T cm^{−1} along the negative *x-*direction (that is it increases by 10^{− 3} T cm^{−1} as one moves in the negative *x*-direction), and it is decreasing in time at the rate of 10^{−3} T s^{−1}. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Side of the square loop, *s* = 12 cm = 0.12 m

Area of the square loop, *A* = 0.12 × 0.12 = 0.0144 m^{2}

Velocity of the loop, *v* = 8 cm/s = 0.08 m/s

Gradient of the magnetic field along negative *x*-direction,

And, rate of decrease of the magnetic field,

Resistance of the loop,

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

Rate of change of the flux due to explicit time variation in field *B *is given as:

Since the rate of change of the flux is the induced *emf*, the total induced *emf *in the loop can be calculated as:

∴Induced current,

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm^{2} with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

Area of the small flat search coil, *A* = 2 cm^{2 }= 2 × 10^{−4} m^{2}

Number of turns on the coil, *N* = 25

Total charge flowing in the coil, *Q* = 7.5 mC = 7.5 × 10^{−3} C

Total resistance of the coil and galvanometer, *R* = 0.50 Ω

Induced current in the coil,

Induced emf is given as:

Where,

= Charge in flux

Combining equations (1) and (2), we get

Initial flux through the coil, = *BA*

Where,

*B* = Magnetic field strength

Final flux through the coil,

Integrating equation (3) on both sides, we have

But total charge,

Hence, the field strength of the magnet is 0.75 T.

Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, *B *= 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

**(a) **Suppose K is open and the rod is moved with a speed of 12 cm s^{−1} in the direction shown. Give the polarity and magnitude of the induced emf.

**(b) **Is there an excess charge built up at the ends of the rods when

K is open? What if K is closed?

**(c) **With K open and the rod moving uniformly, there is *no net force *on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

**(d) **What is the retarding force on the rod when K is closed?

**(e) **How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s^{−1}) when K is closed? How much power is required when K is open?

**(f) **How much power is dissipated as heat in the closed circuit?

What is the source of this power?

**(g) **What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Length of the rod,* l* = 15 cm = 0.15 m

Magnetic field strength,* B* = 0.50 T

Resistance of the closed loop, *R* = 9 mΩ = 9 × 10^{−3} Ω

**(a)** Induced emf = 9 mV; polarity of the induced emf is such that end *P* shows positive while end *Q* shows negative ends.

Speed of the rod, *v* = 12 cm/s = 0.12 m/s

Induced emf is given as:

*e* = *Bvl*

*= *0.5 × 0.12 × 0.15

= 9 × 10^{−3} v

= 9 mV

The polarity of the induced emf is such that end *P* shows positive while end *Q* shows negative ends.

**(b) **Yes; when key K is closed, excess charge is maintained by the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current.

**(c) **Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

**(d) **Retarding force exerted on the rod, *F* = *IBl*

Where,

*I* = Current flowing through the rod

**(e)** 9 mW; no power is expended when key K is open.

Speed of the rod, *v* = 12 cm/s = 0.12 m/s

Hence, power is given as:

When key K is open, no power is expended.

**(f)** 9 mW; power is provided by an external agent.

Power dissipated as heat = *I*^{2} *R*

= (1)^{2} × 9 × 10^{−3}

= 9 mW

The source of this power is an external agent.

**(g)** Zero

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

An air-cored solenoid with length 30 cm, area of cross-section 25 cm^{2} and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^{−3} s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Length of the solenoid, *l* = 30 cm = 0.3 m

Area of cross-section, *A* = 25 cm^{2} = 25 × 10^{−4 }m^{2}

Number of turns on the solenoid, *N* = 500

Current in the solenoid, *I* = 2.5 A

Current flows for time, *t* = 10^{−3} s

Average back emf,

Where,

= Change in flux

= *NAB* … (2)

Where,

*B* = Magnetic field strength

Where,

= Permeability of free space = 4π × 10^{−7} T m A^{−1}

Using equations (2) and (3) in equation (1), we get

Hence, the average back emf induced in the solenoid is 6.5 V.

**(a)** Obtain an expression for the mutual inductance between a long straight wire and a square loop of side *a *as shown in Fig. 6.21.

**(b)** Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, *v *= 10 m/s.

Calculate the induced emf in the loop at the instant when *x *= 0.2 m.

Take *a *= 0.1 m and assume that the loop has a large resistance.

**(a) **Take a small element *dy* in the loop at a distance *y* from the long straight wire (as shown in the given figure).

Magnetic flux associated with element

Where,

*dA* = Area of element *dy *= *a dy*

*B* = Magnetic field at distance *y*

*I* = Current in the wire

= Permeability of free space = 4π × 10^{−7} T m A^{−1}

*y* tends from *x* to .

**(b) **Emf induced in the loop, *e* = *B’av*

Given,

*I *= 50 A

*x* = 0.2 m

*a* = 0.1 m

*v* = 10 m/s

A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass *M *and radius *R*. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

**B **= − B_{0} **k **(*r *≤ *a*; *a *< *R*)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

Line charge per unit length

Where,

*r *= Distance of the point within the wheel

Mass of the wheel = *M*

Radius of the wheel = *R*

Magnetic field,

At distance *r*,themagnetic force is balanced by the centripetal force i.e.,

∴Angular velocity,

**Conclusion**: If you liked NCERT Class 12 Physics Solutions Chapter 6 – Electromagnetic Induction, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 6 – Electromagnetic Induction appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 5 – Magnetism and Matter appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated Class 12 Physics solutions of Chapter 4 i.e Moving Charges and Magnetism. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter:**

Section Name | Topic Name |

5 | Magnetism and Matter |

5.1 | Introduction |

5.2 | The Bar Magnet |

5.3 | Magnetism and Gauss’s Law |

5.4 | The Earth’s Magnetism |

5.5 | Magnetisation and Magnetic Intensity |

5.6 | Magnetic Properties of Materials |

5.7 | Permanent Magnets and Electromagnets |

Answer the following questions regarding earth’s magnetism:

**(a) **A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

**(b) **The angle of dip at a location in southern India is about 18º.

Would you expect a greater or smaller dip angle in Britain?

**(c) **If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

**(d) **In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

**(e) **The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T^{−1} located at its centre. Check the order of magnitude of this number in some way.

**(f ) **Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

**(a) **The three independent quantities conventionally used for specifying earth’s magnetic field are:

(i) Magnetic declination,

(ii) Angle of dip, and

(iii) Horizontal component of earth’s magnetic field

**(b)**The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole

**(c)**It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.

Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

**(d)**If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

**(e)**Magnetic moment, *M* = 8 × 10^{22} J T^{−1}

Radius of earth*, r* = 6.4 × 10^{6} m

Magnetic field strength,

Where,

= Permeability of free space =

This quantity is of the order of magnitude of the observed field on earth.

**(f)**Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

Answer the following questions:

**(a) **The earth’s magnetic field varies from point to point in space.

Does it also change with time? If so, on what time scale does it change appreciably?

**(b)** The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

**(c) **The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?

**(d) **The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?

**(e) **The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

**(f ) **Interstellar space has an extremely weak magnetic field of the order of 10−12 T. Can such a weak field be of any significant consequence? Explain.[**Note: **Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]

**(a) **Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.

**(b)**Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism.

**(c)**Theradioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism.

**(d)**Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.

**(e)**Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.

**(f)**An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^{−2} J. What is the magnitude of magnetic moment of the magnet?

Magnetic field strength, *B* = 0.25 T

Torque on the bar magnet, *T* = 4.5 × 10^{−2} J

Angle between the bar magnet and the external magnetic field,*θ* = 30°

Torque is related to magnetic moment (*M*) as:

*T* = *MB* sin θ

Hence, the magnetic moment of the magnet is 0.36 J T^{−1}_{.}

A short bar magnet of magnetic moment m = 0.32 J T^{−1} is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Moment of the bar magnet, *M* = 0.32 J T^{−1}

External magnetic field, *B* = 0.15 T

**(a)**The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle *θ*, between the bar magnet and the magnetic field is 0°.

Potential energy of the system

**(b)**The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium.

θ = 180°

Potential energy = − *MB* cos *θ*

A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^{−4} m^{2} carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Number of turns in the solenoid, *n *= 800

Area of cross-section, *A* = 2.5 × 10^{−4} m^{2}

Current in the solenoid, *I* = 3.0 A

A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

*M* = *n* *I* *A*

= 800 × 3 × 2.5 × 10^{−4}

= 0.6 J T^{−1}

If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Magnetic field strength,* B *= 0.25 T

Magnetic moment, *M* = 0.6 T^{−1}

The angle *θ*, between the axis of the solenoid and the direction of the applied field is 30°.

Therefore, the torque acting on the solenoid is given as:

A bar magnet of magnetic moment 1.5 J T^{−1} lies aligned with the direction of a uniform magnetic field of 0.22 T.

**(a)** What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

**(b)** What is the torque on the magnet in cases (i) and (ii)?

**(a)**Magnetic moment, *M* = 1.5 J T^{−1}

Magnetic field strength, *B* = 0.22 T

**(i)**Initial angle between the axis and the magnetic field, *θ*_{1} = 0°

Final angle between the axis and the magnetic field, *θ*_{2} = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

**(ii)** Initial angle between the axis and the magnetic field, *θ*_{1} = 0°

Final angle between the axis and the magnetic field, *θ*_{2} = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

**(b)**For case (i):

∴Torque,

For case (ii):

∴Torque,

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^{−4} m^{2}, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^{−2} T is set up at an angle of 30º with the axis of the solenoid?

Number of turns on the solenoid,* n* = 2000

Area of cross-section of the solenoid, *A* = 1.6 × 10^{−4 }m^{2}

Current in the solenoid, *I* = 4 A

**(a)**The magnetic moment along the axis of the solenoid is calculated as:

*M* = *nAI*

= 2000 × 1.6 × 10^{−4} × 4

= 1.28 Am^{2}

**(b)**Magnetic field, *B* = 7.5 × 10^{−2} T

Angle between the magnetic field and the axis of the solenoid, *θ* = 30°

Torque,

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is

A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10^{−2} T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^{−1}. What is the moment of inertia of the coil about its axis of rotation?

Number of turns in the circular coil, *N* = 16

Radius of the coil, *r* = 10 cm = 0.1 m

Cross-section of the coil, *A* = π*r*^{2} = π × (0.1)^{2} m^{2}

Current in the coil, *I* = 0.75 A

Magnetic field strength, *B* = 5.0 × 10^{−2} T

Frequency of oscillations of the coil, *v* = 2.0 s^{−1}

∴Magnetic moment, *M* =* NIA*

= 16 × 0.75 × π × (0.1)^{2}

= 0.377 J T^{−1}

Where,

*I* = Moment of inertia of the coil

Hence, the moment of inertia of the coil about its axis of rotation is

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Horizontal component of earth’s magnetic field, *B*_{H} = 0.35 G

Angle made by the needle with the horizontal plane = Angle of dip =

Earth’s magnetic field strength = *B*

We can relate *B *and *B*_{H}as:

Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

At a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Angle of declination,*θ* = 12°

Angle of dip,

Horizontal component of earth’s magnetic field, *B*_{H} = 0.16 G

Earth’s magnetic field at the given location = *B*

We can relate *B *and *B*_{H}as:

Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

A short bar magnet has a magnetic moment of 0.48 J T^{−1}. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Magnetic moment of the bar magnet, *M* = 0.48 J T^{−1}

**(a****) **Distance, *d* = 10 cm = 0.1 m

The magnetic field at distance *d*, from the centre of the magnet on the axis is given by the relation:

Where,

= Permeability of free space =

The magnetic field is along the S − N direction.

**(b****) **The magnetic field at a distance of 10 cm (i.e., *d* = 0.1 m) on the equatorial line of the magnet is given as:

The magnetic field is along the N − S direction.

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At *null points*, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Earth’s magnetic field at the given place, *H* = 0.36 G

The magnetic field at a distance *d*, on the axis of the magnet is given as:

Where,

= Permeability of free space

*M* = Magnetic moment

The magnetic field at the same distance *d*, on the equatorial line of the magnet is given as:

Total magnetic field,

Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?

The magnetic field on the axis of the magnet at a distance *d*_{1} = 14 cm, can be written as:

Where,

*M *= Magnetic moment

= Permeability of free space

*H* = Horizontal component of the magnetic field at *d*_{1}

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.

Hence, the magnetic field at a distance *d*_{2}, on the equatorial line of the magnet can be written as:

Equating equations (1) and (2), we get:

The new null points will be located 11.1 cm on the normal bisector.

A short bar magnet of magnetic moment 5.25 × 10^{−2} J T^{−1} is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on

(a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Magnetic moment of the bar magnet, *M* = 5.25 × 10^{−2} J T^{−1}

Magnitude of earth’s magnetic field at a place, *H* = 0.42 G = 0.42 × 10^{−4} T

**(****a) **The magnetic field at a distance *R *from the centre of the magnet on the normal bisector is given by the relation:

Where,

= Permeability of free space = 4π × 10^{−7} Tm A^{−1}

When the resultant field is inclined at 45° with earth’s field, *B* = *H*

**(b****) **The magnetic field at a distanced from the centre of the magnet on its axis is given as:

The resultant field is inclined at 45° with earth’s field.

Answer the following questions:

**(a) **Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?

**(b) **Why is diamagnetism, in contrast, almost independent of temperature?

**(c) **If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

**(d) **Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

**(e) **Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

**(f ) **Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?

**(a)** Owing to the random thermal motion of the molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.

**(b)** Each molecule of the diamagnetic material is not a magnetic dipole in itself. Hence, the random thermal motion of the molecules of the diamagnetic material (which is related to the temperature) does not affect the diamagnetism of the material.

**(c)** Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly less than a toroid whose core is empty.

**(d)** The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field.

**(e)** The permeability of a ferromagnetic material is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point. The proof of this fact is based on the boundary conditions of the magnetic fields at the interface of two media.

**(f)** Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.

Answer the following questions:

**(a) **Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

**(b) **The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?

**(c) **‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

**(d)** What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

**(e) **A certain region of space is to be shielded from magnetic fields.

Suggest a method.

The hysteresis curve (*B*–*H* curve) of a ferromagnetic material is shown in the following figure.

**(a) **It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.

**(b)**The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.

**(c)**The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of magnetisation. Hysteresis loops can be used for storing information.

**(d)**Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.

**(e)**A certain region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10° north of east. The magnetic meridian of the place happens to be 10º west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At *neutral points*, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Current in the wire, *I* = 2.5 A

Angle of dip at the given location on earth, = 0°

Earth’s magnetic field, *H* = 0.33 G = 0.33 × 10^{−4} T

The horizontal component of earth’s magnetic field is given as:

*H*_{H} = *H* cos

The magnetic field at the neutral point at a distance *R* from the cable is given by the relation:

Where,

= Permeability of free space =

Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.

A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35º. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Number of horizontal wires in the telephone cable, *n* = 4

Current in each wire, *I* = 1.0 A

Earth’s magnetic field at a location, *H* = 0.39 G = 0.39 × 10^{−4} T

Angle of dip at the location, δ = 35°

Angle of declination, *θ* ∼ 0°

**For a point 4 cm below the cable:**

Distance, *r* = 4 cm = 0.04 m

The horizontal component of earth’s magnetic field can be written as:

*H*_{h} = *H*cos*δ* − *B*

Where,

*B* = Magnetic field at 4 cm due to current *I* in the four wires

= Permeability of free space = 4π × 10^{−7} Tm A^{−1}

= 0.2 × 10^{−4} T = 0.2 G

∴ *H*_{h} = 0.39 cos 35° − 0.2

= 0.39 × 0.819 − 0.2 ≈ 0.12 G

The vertical component of earth’s magnetic field is given as:

*H*_{v} = *H*sin*δ*

= 0.39 sin 35° = 0.22 G

The angle made by the field with its horizontal component is given as:

The resultant field at the point is given as:

**For a point 4 cm above the cable:**

Horizontal component of earth’s magnetic field:

*H*_{h} = *H*cos*δ* + *B*

= 0.39 cos 35° + 0.2 = 0.52 G

Vertical component of earth’s magnetic field:

*H*_{v} = *H*sin*δ*

= 0.39 sin 35° = 0.22 G

Angle, *θ * = 22.9°

And resultant field:

A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

**(a)** Determine the horizontal component of the earth’s magnetic field at the location.

**(b)** The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Number of turns in the circular coil, *N* = 30

Radius of the circular coil, *r* = 12 cm = 0.12 m

Current in the coil, *I* = 0.35 A

Angle of dip, *δ* = 45°

**(a)** The magnetic field due to current *I*, at a distance *r*, is given as:

Where,

= Permeability of free space = 4π × 10^{−7} Tm A^{−1}

= 5.49 × 10^{−5} T

The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as:

*B*_{H} = *B*sin *δ*

= 5.49 × 10^{−5} sin 45° = 3.88 × 10^{−5} T = 0.388 G

**(b)** When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 º, the needle will reverse its original direction. In this case, the needle will point from East to West.

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60º, and one of the fields has a magnitude of 1.2 × 10^{−2} T. If the dipole comes to stable equilibrium at an angle of 15º with this field, what is the magnitude of the other field?

Magnitude of one of the magnetic fields, *B*_{1} = 1.2 × 10^{−2} T

Magnitude of the other magnetic field = *B*_{2}

Angle between the two fields, *θ* = 60°

At stable equilibrium, the angle between the dipole and field *B*_{1}, *θ*_{1} = 15°

Angle between the dipole and field *B*_{2}, *θ*_{2} = *θ* − *θ*_{1 }= 60° − 15° = 45°

At rotational equilibrium, the torques between both the fields must balance each other.

∴Torque due to field *B*_{1} = Torque due to field *B*_{2}

*MB*_{1} sin*θ*_{1} = *MB*_{2} sin*θ*_{2}

Where,

*M* = Magnetic moment of the dipole

Hence, the magnitude of the other magnetic field is 4.39 × 10^{−3} T.

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (*m*_{e}= 9.11 × 10^{−19} C). [**Note: ***Data *in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Energy of an electron beam, *E* = 18 keV = 18 × 10^{3} eV

Charge on an electron, *e* = 1.6 × 10^{−19} C

*E* = 18 × 10^{3} × 1.6 × 10^{−19} J

Magnetic field, *B* = 0.04 G

Mass of an electron, *m*_{e} = 9.11 × 10^{−19} kg

Distance up to which the electron beam travels, *d *= 30 cm = 0.3 m

We can write the kinetic energy of the electron beam as:

The electron beam deflects along a circular path of radius, *r*.

The force due to the magnetic field balances the centripetal force of the path.

Let the up and down deflection of the electron beam be

Where,

*θ* = Angle of declination

Therefore, the up and down deflection of the beam is 3.9 mm.

A sample of paramagnetic salt contains 2.0 × 10^{24} atomic dipoles each of dipole moment 1.5 × 10^{−23 }J T^{−1}. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

Number of atomic dipoles, *n* = 2.0 × 10^{24}

Dipole moment of each atomic dipole, *M* = 1.5 × 10^{−23} J T^{−1}

When the magnetic field, *B*_{1} = 0.64 T

The sample is cooled to a temperature, *T*_{1} = 4.2°K

Total dipole moment of the atomic dipole, *M*_{tot} = *n* × *M*

= 2 × 10^{24} × 1.5 × 10^{−23}

= 30 J T^{−1}

Magnetic saturation is achieved at 15%.

Hence, effective dipole moment,

When the magnetic field, *B*_{2} = 0.98 T

Temperature, *T*_{2} = 2.8°K

Its total dipole moment = *M*_{2}

According to Curie’s law, we have the ratio of two magnetic dipoles as:

Therefore, is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.

A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field **B **in the core for a magnetising current of 1.2 A?

Mean radius of a Rowland ring, *r* = 15 cm = 0.15 m

Number of turns on a ferromagnetic core, *N* = 3500

Relative permeability of the core material,

Magnetising current, *I* = 1.2 A

The magnetic field is given by the relation:

*B *

Where,

μ_{0} = Permeability of free space = 4π × 10^{−7} Tm A^{−1}

Therefore, the magnetic field in the core is 4.48 T.

The magnetic moment vectors μ_{s}and μ_{l}associated with the intrinsic spin angular momentum **S **and orbital angular momentum **l**, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:

μ_{s}= –(*e*/*m*) **S**,

μ_{l}* = *–(*e*/2*m*)**l**

Which of these relations is in accordance with the result expected *classically*? Outline the derivation of the classical result.

The magnetic moment associated with the orbital angular momentum is valid with the classical mechanics.

The magnetic moment associated with the orbital angular momentum is given as

For current *i *and area of cross-section *A*, we have the relation:

Magnetic moment

………………………….(1)

Where,

*e*= Charge of the electron

*r*= Radius of the circular orbit

*T*= Time taken to complete one rotation around the circular orbit of radius *r*

Orbital angular momentum, *l*=* mvr*

…………………………….(2)

Where,

*m*= Mass of the electron

*v*= Velocity of the electron

r= Radius of the circular orbit

Dividing equation (1) by equation (2), we get:

**Conclusion**: If you liked NCERT Class 12 Physics Solutions Chapter 5 – Magnetism and Matter, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 5 – Magnetism and Matter appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 4 – Moving Charges And Magnetism appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated Class 12 Physics solutions of Chapter 3 i.e Current Electricity. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism:**

Section Name | Topic Name |

4 | Moving Charges and Magnetism |

4.1 | Introduction |

4.2 | Magnetic Force |

4.3 | Motion in a Magnetic Field |

4.4 | Motion in Combined Electric and Magnetic Fields |

4.5 | Magnetic Field due to a Current Element, Biot-Savart Law |

4.6 | Magnetic Field on the Axis of a Circular Current Loop |

4.7 | Ampere’s Circuital Law |

4.8 | The Solenoid and the Toroid |

4.9 | Force between Two Parallel Currents, the Ampere |

4.10 | Torque on Current Loop, Magnetic Dipole |

4.11 | Torque on Current Loop, Magnetic Dipole |

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field **B **at the centre of the coil?

Number of turns on the circular coil, *n* = 100

Radius of each turn,* r* = 8.0 cm = 0.08 m

Current flowing in the coil, *I* = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given by the relation,

Where,

= Permeability of free space

= 4π × 10^{–7} T m A^{–1}

Hence, the magnitude of the magnetic field is 3.14 × 10^{–4} T.

A long straight wire carries a current of 35 A. What is the magnitude of the field **B **at a point 20 cm from the wire?

Current in the wire, *I* = 35 A

Distance of a point from the wire, *r* = 20 cm = 0.2 m

Magnitude of the magnetic field at this point is given as:

*B*

Where,

= Permeability of free space = 4π × 10^{–7} T m A^{–1}

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10^{–5} T.

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of **B **at a point 2.5 m east of the wire.

Current in the wire, *I* = 50 A

A point is 2.5 m away from the East of the wire.

Magnitude of the distance of the point from the wire, *r* = 2.5 m.

Magnitude of the magnetic field at that point is given by the relation, *B*

Where,

_{ =} Permeability of free space = 4π × 10^{–7} T m A^{–1}

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Current in the power line, *I* = 90 A

Point is located below the power line at distance, *r* = 1.5 m

Hence, magnetic field at that point is given by the relation,

Where,

_{ =} Permeability of free space = 4π × 10^{–7} T m A^{–1}

The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Current in the wire, *I *= 8 A

Magnitude of the uniform magnetic field, *B* = 0.15 T

Angle between the wire and magnetic field, *θ* = 30°.

Magnetic force per unit length on the wire is given as:

*f* = *BI* sin*θ*

= 0.15 × 8 ×1 × sin30°

= 0.6 N m^{–1}

Hence, the magnetic force per unit length on the wire is 0.6 N m^{–1}.

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Length of the wire, *l* = 3 cm = 0.03 m

Current flowing in the wire, *I* = 10 A

Magnetic field, *B* = 0.27 T

Angle between the current and magnetic field, *θ* = 90°

Magnetic force exerted on the wire is given as:

*F* = *BIl*sin*θ*

= 0.27 × 10 × 0.03 sin90°

= 8.1 × 10^{–2} N

Hence, the magnetic force on the wire is 8.1 × 10^{–2} N. The direction of the force can be obtained from Fleming’s left hand rule.

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Current flowing in wire A, *I*_{A} = 8.0 A

Current flowing in wire B, *I*_{B} = 5.0 A

Distance between the two wires, *r* = 4.0 cm = 0.04 m

Length of a section of wire A,* l* = 10 cm = 0.1 m

Force exerted on length *l* due to the magnetic field is given as:

Where,

= Permeability of free space = 4π × 10^{–7} T m A^{–1}

The magnitude of force is 2 × 10^{–5} N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of **B **inside the solenoid near its centre.

Length of the solenoid, *l* = 80 cm = 0.8 m

There are five layers of windings of 400 turns each on the solenoid.

Total number of turns on the solenoid, *N* = 5 × 400 = 2000

Diameter of the solenoid, *D* = 1.8 cm = 0.018 m

Current carried by the solenoid, *I* = 8.0 A

Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

Where,

= Permeability of free space = 4π × 10^{–7} T m A^{–1}

Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 × 10^{–2} T.

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Length of a side of the square coil, *l* = 10 cm = 0.1 m

Current flowing in the coil, *I* = 12 A

Number of turns on the coil, *n* = 20

Angle made by the plane of the coil with magnetic field, *θ* = 30°

Strength of magnetic field, *B* = 0.80 T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

τ = *n* *BIA* sin*θ*

Where,

*A* = Area of the square coil

*l × l* = 0.1 × 0.1 = 0.01 m^{2}

∴ τ = 20 × 0.8 × 12 × 0.01 × sin30°

= 0.96 N m

Hence, the magnitude of the torque experienced by the coil is 0.96 N m.

Two moving coil meters, M_{1} and M_{2} have the following particulars:

*R*_{1} = 10 Ω, *N*_{1} = 30,

*A*_{1} = 3.6 × 10^{–3} m^{2}*, **B*_{1} = 0.25 T

*R*_{2} = 14 Ω, *N*_{2} = 42,

*A*_{2} = 1.8 × 10^{–3} m^{2}, *B*_{2} = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}.

For moving coil meter M_{1}:

Resistance, *R*_{1} = 10 Ω

Number of turns, *N*_{1} = 30

Area of cross-section, *A*_{1} = 3.6 × 10^{–3} m^{2}

Magnetic field strength, *B*_{1} = 0.25 T

Spring constant *K*_{1} = *K*

For moving coil meter M_{2}:

Resistance, *R*_{2} = 14 Ω

Number of turns, *N*_{2} = 42

Area of cross-section, *A*_{2} = 1.8 × 10^{–3} m^{2}

Magnetic field strength, *B*_{2} = 0.50 T

Spring constant, *K*_{2} = *K*

**(a)** Current sensitivity of M_{1} is given as:

And, current sensitivity of M_{2} is given as:

Ratio

Hence, the ratio of current sensitivity of M_{2} to M_{1} is 1.4.

**(b)** Voltage sensitivity for M_{2} is given as:

And, voltage sensitivity for M_{1} is given as:

Vs1=N1B1A1K1R1

Ratio

Vs2Vs1=N2B2A2K1R1K2R2N1B1A1

Hence, the ratio of voltage sensitivity of M_{2} to M_{1} is 1.

In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^{–4} T) is maintained. An electron is shot into the field with a speed of 4.8 × 10^{6} m s^{–1} normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (*e *= 1.6 × 10^{–19} C, *m*_{e}= 9.1×10^{–31} kg)

Magnetic field strength, *B* = 6.5 G = 6.5 × 10^{–4} T

Speed of the electron, *v* = 4.8 × 10^{6} m/s

Charge on the electron, *e* = 1.6 × 10^{–19} C

Mass of the electron, *m*_{e} = 9.1 × 10^{–31} kg

Angle between the shot electron and magnetic field, *θ* = 90°

Magnetic force exerted on the electron in the magnetic field is given as:

*F* = *evB *sin*θ*

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius *r*.

Hence, centripetal force exerted on the electron,

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

Hence, the radius of the circular orbit of the electron is 4.2 cm.

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Magnetic field strength, *B* = 6.5 × 10^{−4} T

Charge of the electron, *e* = 1.6 × 10^{−19} C

Mass of the electron, *m*_{e} = 9.1 × 10^{−31} kg

Velocity of the electron, *v* = 4.8 × 10^{6} m/s

Radius of the orbit, *r* = 4.2 cm = 0.042 m

Frequency of revolution of the electron = *ν*

Angular frequency of the electron = *ω* = 2π*ν*

Velocity of the electron is related to the angular frequency as:

*v* = *rω*

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:

This expression for frequency is independent of the speed of the electron.

On substituting the known values in this expression, we get the frequency as:

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

**(a)** A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

**(b)** Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

**(a)** Number of turns on the circular coil, *n* = 30

Radius of the coil, *r* = 8.0 cm = 0.08 m

Area of the coil

Current flowing in the coil, *I* = 6.0 A

Magnetic field strength, *B* = 1 T

Angle between the field lines and normal with the coil surface,

*θ* = 60°

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

τ = *n* *IBA* sin*θ* … (*i*)

= 30 × 6 × 1 × 0.0201 × sin60°

= 3.133 N m

**(b)** It can be inferred from relation (*i*) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Radius of coil X, *r*_{1} = 16 cm = 0.16 m

Radius of coil Y,* r*_{2} = 10 cm = 0.1 m

Number of turns of on coil X, *n*_{1} = 20

Number of turns of on coil Y, *n*_{2} = 25

Current in coil X, *I*_{1} = 16 A

Current in coil Y, *I*_{2} = 18 A

Magnetic field due to coil X at their centre is given by the relation,

Where,

= Permeability of free space =

Magnetic field due to coil Y at their centre is given by the relation,

Hence, net magnetic field can be obtained as:

A magnetic field of 100 G (1 G = 10^{−4} T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^{−3} m^{2}. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^{−1}. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic

Magnetic field strength,* B* = 100 G = 100 × 10^{−4} T

Number of turns per unit length, *n* = 1000 turns m^{−1}

Current flowing in the coil, *I* = 15 A

Permeability of free space, =

Magnetic field is given by the relation,

If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.

For a circular coil of radius *R *and *N *turns carrying current *I*, the magnitude of the magnetic field at a point on its axis at a distance *x *from its centre is given by,

**(a)** Show that this reduces to the familiar result for field at the centre of the coil.

**(b)** Consider two parallel co-axial circular coils of equal radius *R*, and number of turns *N*, carrying equal currents in the same direction, and separated by a distance *R*. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to *R*, and is given by,

, approximately.[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as *Helmholtz coils*.]

Radius of circular coil = *R*

Number of turns on the coil = *N*

Current in the coil = *I*

Magnetic field at a point on its axis at distance *x* is given by the relation,

Where,

= Permeability of free space

**(a) **If the magnetic field at the centre of the coil is considered, then *x* = 0.

This is the familiar result for magnetic field at the centre of the coil.

**(b) **Radii of two parallel co-axial circular coils = *R*

Number of turns on each coil = *N*

Current in both coils = *I*

Distance between both the coils = *R*

Let us consider point Q at distance *d* from the centre.

Then, one coil is at a distance of from point Q.

Magnetic field at point Q is given as:

Also, the other coil is at a distance of from point Q.

Magnetic field due to this coil is given as:

Total magnetic field,

Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.

A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Inner radius of the toroid, *r*_{1} = 25 cm = 0.25 m

Outer radius of the toroid, *r*_{2} = 26 cm = 0.26 m

Number of turns on the coil,* N* = 3500

Current in the coil, *I* = 11 A

**(a) **Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.

**(b) **Magnetic field inside the core of a toroid is given by the relation,

*B* =

Where,

= Permeability of free space =

*l* = length of toroid

**(c) **Magnetic field in the empty space surrounded by the toroid is zero.

Answer the following questions:

**(a)** A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

**(b)** A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

**(c)** An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

**(a) **The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.

**(b) **Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.

**(c) **An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.

An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.

Magnetic field strength, *B* = 0.15 T

Charge on the electron, *e* = 1.6 × 10^{−19 }C

Mass of the electron, *m* = 9.1 × 10^{−31 }kg

Potential difference, *V* = 2.0 kV = 2 × 10^{3} V

Thus, kinetic energy of the electron = *eV*

Where,

*v *= velocity of the electron

**(a) **Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius *r*.

Magnetic force on the electron is given by the relation,

*B ev*

Centripetal force

From equations (1) and (2), we get

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

**(b)** When the field makes an angle *θ* of 30° with initial velocity, the initial velocity will be,

From equation (2), we can write the expression for new radius as:

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.

A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^{5} V m^{−1}, make a simple guess as to what the beam contains. Why is the answer not unique?

Magnetic field, *B* = 0.75 T

Accelerating voltage, *V* = 15 kV = 15 × 10^{3} V

Electrostatic field, *E* = 9 × 10^{5} V m^{−1}

Mass of the electron = *m*

Charge of the electron = *e*

Velocity of the electron = *v*

Kinetic energy of the electron = *eV*

Since the particle remains undeflected by electric and magnetic fields, we can infer that the force on the charged particle due to electric field is balancing the force on the charged particle due to magnetic field.

Putting equation (2) in equation (1), we get

This value of specific charge *e*/*m* is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He^{++,} Li^{++}, etc.

A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

**(a)** What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

**(b)** What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s^{−2}.

Length of the rod,* l* = 0.45 m

Mass suspended by the wires, *m* = 60 g = 60 × 10^{−3} kg

Acceleration due to gravity, g = 9.8 m/s^{2}

Current in the rod flowing through the wire, *I* = 5 A

**(a) **Magnetic field (*B*) is equal and opposite to the weight of the wire i.e.,

A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming’s left hand rule gives an upward magnetic force.

**(b) **If the direction of the current is revered, then the force due to magnetic field and the weight of the wire acts in a vertically downward direction.

∴Total tension in the wire = *BIl* + *m*g

The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Current in both wires, *I* = 300 A

Distance between the wires, *r *= 1.5 cm = 0.015 m

Length of the two wires, *l* = 70 cm = 0.7 m

Force between the two wires is given by the relation,

Where,

= Permeability of free space =

Since the direction of the current in the wires is opposite, a repulsive force exists between them.

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

**(a)** the wire intersects the axis,

**(b)** the wire is turned from N-S to northeast-northwest direction,

**(c)** the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Magnetic field strength, *B* = 1.5 T

Radius of the cylindrical region, *r* = 10 cm = 0.1 m

Current in the wire passing through the cylindrical region, *I* = 7 A

**(a) **If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region.

Thus, *l* = 2*r* = 0.2 m

Angle between magnetic field and current, *θ* = 90°

Magnetic force acting on the wire is given by the relation,

*F* = *BIl* sin *θ*

= 1.5 × 7 × 0.2 × sin 90°

= 2.1 N

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

**(b)** New length of the wire after turning it to the Northeast-Northwest direction can be given as: :

Angle between magnetic field and current, *θ* = 45°

Force on the wire,

*F* = *BIl*_{1} sin *θ*

Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle*θ*because *l* sin*θ* is fixed.

**(c) **The wire is lowered from the axis by distance, *d* = 6.0 cm

Suppose wire is passing perpendicularly to the axis of cylindrical magnetic field then lowering 6 cm means displacing the wire 6 cm from its initial position towards to end of cross sectional area.

Thus the length of wire in magnetic field will be 16 cm as AB= *L *=2*x* =16 cm

Now the force,

*F *= *iLB *sin90° as the wire will be perpendicular to the magnetic field.

F= 7 × 0.16 × 1.5 =1.68 N

The direction will be given by right hand curl rule or screw rule i.e. vertically downwards.

A uniform magnetic field of 3000 G is established along the positive *z*-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Magnetic field strength, *B* = 3000 G = 3000 × 10^{−4} T = 0.3 T

Length of the rectangular loop,* l* = 10 cm

Width of the rectangular loop, *b* = 5 cm

Area of the loop,

*A* = *l* × *b* = 10 × 5 = 50 cm^{2 }= 50 × 10^{−4} m^{2}

Current in the loop, *I* = 12 A

Now, taking the anti-clockwise direction of the current as positive and vise-versa:

**(a) **Torque,

From the given figure, it can be observed that *A *is normal to the *y*–*z* plane and *B* is directed along the *z*-axis.

The torque is N m along the negative *y*-direction. The force on the loop is zero because the angle between *A* and *B* is zero.

**(b)** This case is similar to case (a). Hence, the answer is the same as (a).

**(c)** Torque

From the given figure, it can be observed that *A *is normal to the *x*–*z* plane and *B* is directed along the *z*-axis.

The torque is N m along the negative *x *direction and the force is zero.

**(d)** Magnitude of torque is given as:

Torque is N m at an angle of 240° with positive *x *direction. The force is zero.

**(e)** Torque

Hence, the torque is zero. The force is also zero.

**(f)** Torque

Hence, the torque is zero. The force is also zero.

In case (e), the direction of and is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.

Whereas, in case (f), the direction of and is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the

**(a) **total torque on the coil,

**(b)** total force on the coil,

**(c) **average force on each electron in the coil due to the magnetic field?

(The coil is made of copper wire of cross-sectional area 10^{−5} m^{2}, and the free electron density in copper is given to be about 10^{29} m^{−3}.)

Number of turns on the circular coil,* n* = 20

Radius of the coil,* r* = 10 cm = 0.1 m

Magnetic field strength, *B* = 0.10 T

Current in the coil, *I* = 5.0 A

**(a) **The total torque on the coil is zero because the field is uniform.

**(b) **The total force on the coil is zero because the field is uniform.

**(c) **Cross-sectional area of copper coil, *A* = 10^{−5} m^{2}

Number of free electrons per cubic meter in copper, *N* = 10^{29 }/m^{3}

Charge on the electron, *e* = 1.6 × 10^{−19} C

Magnetic force, *F* = *Bev*_{d}

Where,

*v*_{d}= Drift velocity of electrons

Hence, the average force on each electron is

A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? *g *= 9.8 m s^{−2}

Length of the solenoid, *L* = 60 cm = 0.6 m

Radius of the solenoid,* r* = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each.

Total number of turns, *n* = 3 × 300 = 900

Length of the wire, *l* = 2 cm = 0.02 m

Mass of the wire, *m* = 2.5 g = 2.5 × 10^{−3} kg

Current flowing through the wire, *i* = 6 A

Acceleration due to gravity, g = 9.8 m/s^{2}

Magnetic field produced inside the solenoid,

Where,

= Permeability of free space =

*I *= Current flowing through the windings of the solenoid

Magnetic force is given by the relation,

Also, the force on the wire is equal to the weight of the wire.

Hence, the current flowing through the solenoid is 108 A.

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Resistance of the galvanometer coil, *G* = 12 Ω

Current for which there is full scale deflection, = 3 mA = 3 × 10^{−3} A

Range of the voltmeter is 0, which needs to be converted to 18 V.

*V* = 18 V

Let a resistor of resistance* R* be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:

Hence, a resistor of resistance is to be connected in series with the galvanometer.

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Resistance of the galvanometer coil, *G* = 15 Ω

Current for which the galvanometer shows full scale deflection,

= 4 mA = 4 × 10^{−3} A

Range of the ammeter is 0, which needs to be converted to 6 A.

Current, *I* = 6 A

A shunt resistor of resistance *S* is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of *S *is given as:

Hence, a shunt resistor is to be connected in parallel with the galvanometer.

**Conclusion**: If you liked NCERT Class 12 Physics Solutions Chapter 4 – Moving Charges And Magnetism, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 4 – Moving Charges And Magnetism appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 3 – Current Electricity appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated Class 12 Physics solutions of Chapter 2 i.e Electrostatic Potential And Capacitance. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity:**

Section Name | Topic Name |

3 | Current Electricity |

3.1 | Introduction |

3.2 | Electric Current |

3.3 | Electric Currents in Conductors |

3.4 | Ohm’s law |

3.5 | Drift of Electrons and the Origin of Resistivity |

3.6 | Limitations of Ohm’s Law |

3.7 | Resistivity of Various Materials |

3.8 | Temperature Dependence of Resistivity |

3.9 | Electrical Energy, Power |

3.10 | Combination of Resistors — Series and Parallel |

3.11 | Cells, emf, Internal Resistance |

3.12 | Cells in Series and in Parallel |

3.13 | Kirchhoff’s Rules |

3.14 | Wheatstone Bridge |

3.15 | Meter Bridge |

3.16 | Potentiometer |

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?

Emf of the battery, *E* = 12 V

Internal resistance of the battery, *r* = 0.4 Ω

Maximum current drawn from the battery = *I*

According to Ohm’s law,

The maximum current drawn from the given battery is 30 A.

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Emf of the battery, *E* = 10 V

Internal resistance of the battery, *r* = 3 Ω

Current in the circuit, *I* = 0.5 A

Resistance of the resistor = *R*

The relation for current using Ohm’s law is,

Terminal voltage of the resistor = *V*

According to Ohm’s law,

*V* = *IR*

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.

**(a)** Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

**(b)** If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

**(a)** Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

**(b)** Current flowing through the circuit = *I*

Emf of the battery, *E* = 12 V

Total resistance of the circuit, *R *= 6 Ω

The relation for current using Ohm’s law is,

Potential drop across 1 Ω resistor = *V*_{1}

From Ohm’s law, the value of *V*_{1 }can be obtained as

*V*_{1} = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = *V*_{2}

Again, from Ohm’s law, the value of *V*_{2 }can be obtained as

*V*_{2} = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = *V*_{3}

Again, from Ohm’s law, the value of *V*_{3 }can be obtained as

*V*_{3} = 2 × 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.

**(a) **Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

**(b) **If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

**(a)** There are three resistors of resistances,

*R*_{1} = 2 Ω, *R*_{2} = 4 Ω, and *R*_{3} = 5 Ω

They are connected in parallel. Hence, total resistance (*R*) of the combination is given by,

Therefore, total resistance of the combination is.

**(b)** Emf of the battery, *V* = 20 V

Current (*I*_{1}) flowing through resistor *R*_{1} is given by,

Current (*I*_{2}) flowing through resistor *R*_{2} is given by,

Current (*I*_{3}) flowing through resistor *R*_{3} is given by,

Total current, *I* = *I*_{1} + *I*_{2} + *I*_{3} = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.

At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is

Room temperature, *T* = 27°C

Resistance of the heating element at *T*, *R* = 100 Ω

Let *T*_{1} is the increased temperature of the filament.

Resistance of the heating element at *T*_{1}, *R*_{1} = 117 Ω

Temperature co-efficient of the material of the filament,

Therefore, at 1027°C, the resistance of the element is 117Ω.

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^{−7 }m^{2}, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Length of the wire, *l* =15 m

Area of cross-section of the wire, *a* = 6.0 × 10^{−7 }m^{2}

Resistance of the material of the wire, *R* = 5.0 Ω

Resistivity of the material of the wire = *ρ*

Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10^{−7 }Ω m.

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Temperature, *T*_{1} = 27.5°C

Resistance of the silver wire at *T*_{1}, *R*_{1} = 2.1 Ω

Temperature, *T*_{2} = 100°C

Resistance of the silver wire at *T*_{2}, *R*_{2} = 2.7 Ω

Temperature coefficient of silver = *α*

It is related with temperature and resistance as

Therefore, the temperature coefficient of silver is 0.0039°C^{−1}.

Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^{−4 }°C ^{−1}.

Supply voltage,* V* = 230 V

Initial current drawn, *I*_{1} = 3.2 A

Initial resistance = *R*_{1}, which is given by the relation,

Steady state value of the current, *I*_{2} = 2.8 A

Resistance at the steady state = *R*_{2}, which is given as

Temperature co-efficient of nichrome, *α* = 1.70 × 10^{−4 }°C ^{−1}

Initial temperature of nichrome, *T*_{1}= 27.0°C

Study state temperature reached by nichrome = *T*_{2}

*T*_{2 }can be obtained by the relation for *α*,

Therefore, the steady temperature of the heating element is 867.5°C

Determine the current in each branch of the network shown in fig 3.30:

Current flowing through various branches of the circuit is represented in the given figure.

*I*_{1} = Current flowing through the outer circuit

*I*_{2} = Current flowing through branch AB

*I*_{3} = Current flowing through branch AD

*I*_{2} − *I*_{4} = Current flowing through branch BC

*I*_{3} + *I*_{4} = Current flowing through branch CD

*I*_{4} = Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

10*I*_{2} + 5*I*_{4} − 5*I*_{3} = 0

2*I*_{2} + *I*_{4} −*I*_{3} = 0

*I*_{3} = 2*I*_{2} + *I*_{4} … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(*I*_{2} − *I*_{4}) − 10(*I*_{3} +* I*_{4}) − 5*I*_{4 }= 0

5*I*_{2} + 5*I*_{4} − 10*I*_{3} − 10*I*_{4 }− 5*I*_{4 }= 0

5*I*_{2} − 10*I*_{3} − 20*I*_{4 }= 0

*I*_{2} = 2*I*_{3} + 4*I*_{4 }… (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (*I*_{1}) + 10(*I*_{2}) + 5(*I*_{2} −* I*_{4}) = 0

10 = 15*I*_{2} + 10*I*_{1 }− 5*I*_{4}

3*I*_{2} + 2*I*_{1 }− *I*_{4} = 2 … (3)

From equations (1) and (2), we obtain

*I*_{3} = 2(2*I*_{3 }+ 4*I*_{4}) +* I*_{4}

*I*_{3} = 4*I*_{3 }+ 8*I*_{4} +* I*_{4}

− 3*I*_{3} = 9*I*_{4}

− 3*I*_{4} = + *I*_{3 }… (4)

Putting equation (4) in equation (1), we obtain

*I*_{3} = 2*I*_{2 }+ *I*_{4}

− 4*I*_{4} = 2*I*_{2}

*I*_{2} = − 2*I*_{4} … (5)

It is evident from the given figure that,

*I*_{1} = *I*_{3 }+ *I*_{2} … (6)

Putting equation (6) in equation (1), we obtain

3*I*_{2} +2(*I*_{3 }+ *I*_{2}) −* I*_{4} = 2

5*I*_{2} + 2*I*_{3 }− *I*_{4} = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2* I*_{4}) + 2(− 3 *I*_{4})_{ −} *I*_{4} = 2

− 10*I*_{4} − 6*I*_{4 }− *I*_{4} = 2

17*I*_{4} = − 2

Equation (4) reduces to

*I*_{3} = − 3(*I*_{4})

Therefore, current in branch

In branch BC =

In branch CD =

In branch AD

In branch BD =

Total current =

**(a)** In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end *A*, when the resistor *Y *is of 12.5 Ω. Determine the resistance of *X*. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

**(b)** Determine the balance point of the bridge above if *X *and *Y *are interchanged.

**(c)** What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

A metre bridge with resistors *X* and *Y* is represented in the given figure.

**(a)** Balance point from end A, *l*_{1} = 39.5 cm

Resistance of the resistor *Y* = 12.5 Ω

Condition for the balance is given as,

Therefore, the resistance of resistor *X* is 8.2 Ω.

The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

**(b) **If *X *and *Y* are interchanged, then *l*_{1} and 100−*l*_{1} get interchanged.

The balance point of the bridge will be 100−*l*_{1} from A.

100−*l*_{1 }= 100 − 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.

**(c) **When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Emf of the storage battery, *E* = 8.0 V

Internal resistance of the battery, *r* = 0.5 Ω

DC supply voltage, *V* = 120 V

Resistance of the resistor, *R* = 15.5 Ω

Effective voltage in the circuit =* V*^{1}

*R *is connected to the storage battery in series. Hence, it can be written as

*V*^{1} = *V* − *E*

*V*^{1 }= 120 − 8 = 112 V

Current flowing in the circuit = *I*, which is given by the relation,

Voltage across resistor *R* given by the product, *IR *= 7 × 15.5 = 108.5 V

DC supply voltage = Terminal voltage of battery + Voltage drop across *R*

Terminal voltage of battery = 120 − 108.5 = 11.5 V

A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Emf of the cell, *E*_{1} = 1.25 V

Balance point of the potentiometer, *l*_{1}= 35 cm

The cell is replaced by another cell of emf *E*_{2}.

New balance point of the potentiometer, *l*_{2} = 63 cm

Therefore, emf of the second cell is 2.25V.

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10^{28} m^{−3}. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^{−6 }m^{2} and it is carrying a current of 3.0 A.

Number density of free electrons in a copper conductor, *n* = 8.5 × 10^{28} m^{−3} Length of the copper wire, *l* = 3.0 m

Area of cross-section of the wire, *A* = 2.0 × 10^{−6 }m^{2}

Current carried by the wire, *I* = 3.0 A, which is given by the relation,

*I *= *nA*e*V*_{d}

Where,

e = Electric charge = 1.6 × 10^{−19} C

*V*_{d} = Drift velocity

Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 10^{4} s.

The earth’s surface has a negative surface charge density of 10^{−9} C m^{−2}. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10^{6} m.)

Surface charge density of the earth, *σ* = 10^{−9} C m^{−2}

Current over the entire globe, *I* = 1800 A

Radius of the earth, *r* = 6.37 × 10^{6} m

Surface area of the earth,

*A* = 4π*r*^{2}

= 4π × (6.37 × 10^{6})^{2}

= 5.09 × 10^{14} m^{2}

Charge on the earth surface,

*q *= *σ* × *A*

= 10^{−9} × 5.09 × 10^{14}

= 5.09 × 10^{5 }C

Time taken to neutralize the earth’s surface =* t*

Current,

Therefore, the time taken to neutralize the earth’s surface is 282.77 s.

**(a)** Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

**(b)** A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

**(a) **Number of secondary cells, *n* = 6

Emf of each secondary cell, *E* = 2.0 V

Internal resistance of each cell, *r* = 0.015 Ω

series resistor is connected to the combination of cells.

Resistance of the resistor, *R* = 8.5 Ω

Current drawn from the supply = *I*, which is given by the relation,

Terminal voltage,* V* = *IR* = 1.39 × 8.5 = 11.87 A

Therefore, the current drawn from the supply is 1.39 A and terminal voltage is

11.87 A.

**(b) **After a long use, emf of the secondary cell, *E* = 1.9 V

Internal resistance of the cell, *r* = 380 Ω

Hence, maximum current

Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (*ρ*_{Al} = 2.63 × 10^{−8} Ω m, *ρ*_{Cu} = 1.72 × 10^{−8} Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Resistivity of aluminium, *ρ*_{Al} = 2.63 × 10^{−8} Ω m

Relative density of aluminium, *d*_{1} = 2.7

Let *l*_{1} be the length of aluminium wire and *m*_{1} be its mass.

Resistance of the aluminium wire = *R*_{1}

Area of cross-section of the aluminium wire =* A*_{1}

Resistivity of copper, *ρ*_{Cu} = 1.72 × 10^{−8} Ω m

Relative density of copper, *d*_{2} = 8.9

Let *l*_{2} be the length of copper wire and *m*_{2} be its mass.

Resistance of the copper wire =* R*_{2}

Area of cross-section of the copper wire =* A*_{2}

The two relations can be written as

It is given that,

And,

Mass of the aluminium wire,

*m*_{1} = Volume × Density

= *A*_{1}*l*_{1} × *d*_{1 }= *A*_{1} *l*_{1}*d*_{1} … (3)

Mass of the copper wire,

*m*_{2} = Volume × Density

= *A*_{2}*l*_{2} × *d*_{2 }= *A*_{2} *l*_{2}*d*_{2} … (4)

Dividing equation (3) by equation (4), we obtain

It can be inferred from this ratio that *m*_{1} is less than *m*_{2}. Hence, aluminium is lighter than copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.

What conclusion can you draw from the following observations on a resistor made of alloy manganin?

CurrentA | VoltageV | CurrentA | VoltageV |

0.2 | 3.94 | 3.0 | 59.2 |

0.4 | 7.87 | 4.0 | 78.8 |

0.6 | 11.8 | 5.0 | 98.6 |

0.8 | 15.7 | 6.0 | 118.5 |

1.0 | 19.7 | 7.0 | 138.2 |

2.0 | 39.4 | 8.0 | 158.0 |

It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.

Answer the following questions:

**(a) **A steady current flows in a metallic conductor of non-uniform cross- section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

**(b)** Is Ohm’s law universally applicable for all conducting elements?

If not, give examples of elements which do not obey Ohm’s law.

**(c) **A low voltage supply from which one needs high currents must have very low internal resistance. Why?

**(d) **A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

**(a) **When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.

**(b) **No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.

**(c) **According to Ohm’s law, the relation for the potential is *V *=* IR*

Voltage (*V*) is directly proportional to current (*I*).

*R* is the internal resistance of the source.

If *V* is low, then *R* must be very low, so that high current can be drawn from the source.

**(d) **In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.

Choose the correct alternative:

**(a) **Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

**(b) **Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

**(c) **The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.

**(d) **The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10^{22}/10^{3}).

**(a) **Alloys of metals usually have greater resistivity than that of their constituent metals.

**(b) **Alloys usually have lower temperature coefficients of resistance than pure metals.

**(c) **The resistivity of the alloy, manganin, is nearly independent of increase of temperature.

**(d) **The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 10^{22}.

**(a)** Given *n *resistors each of resistance *R*, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

**(b)** Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

**(c)** Determine the equivalent resistance of networks shown in Fig. 3.31.

**(a) **Total number of resistors = *n*

Resistance of each resistor = *R*

**(i)** When *n* resistors are connected in series, effective resistance *R*_{1}is the maximum, given by the product *nR*.

Hence, maximum resistance of the combination, *R*_{1 }= *nR*

**(ii)** When *n* resistors are connected in parallel, the effective resistance (*R*_{2}) is the minimum, given by the ratio.

Hence, minimum resistance of the combination, *R*_{2} =

**(iii)** The ratio of the maximum to the minimum resistance is,

**(b) **The resistance of the given resistors is,

*R*_{1} = 1 Ω, *R*_{2} = 2 Ω, *R*_{3} = 3 Ω2

- Equivalent resistance,

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

- Equivalent resistance,

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

**(iii)** Equivalent resistance, *R*^{’} = 6 Ω

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by the sum,

*R*^{’} = 1 + 2 + 3 = 6 Ω

**(iv)** Equivalent resistance,

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by,

**(c) (a)** It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1+1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Therefore, the circuit can be redrawn as

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four loops. Hence, equivalent resistance (*R*^{’}) of each loop is given by,

The circuit reduces to,

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is

**(b)** It can be observed from the given circuit that five resistors of resistance *R* each are connected in series.

Hence, equivalent resistance of the circuit = *R *+ *R* +* R* + *R* + *R*

= 5 *R*

*2*

Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.

The resistance of each resistor connected in the given circuit, *R* = 1 Ω

Equivalent resistance of the given circuit = *R*^{’}

The network is infinite. Hence, equivalent resistance is given by the relation,

Negative value of *R*^{’ }cannot be accepted. Hence, equivalent resistance,

Internal resistance of the circuit, *r* = 0.5 Ω

Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

Supply voltage, *V* = 12 V

According to Ohm’s Law, current drawn from the source is given by the ratio, = 3.72 A

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf *ε* and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

**(a) **What is the value *ε* ?

**(b) **What purpose does the high resistance of 600 kΩ have?

**(c) **Is the balance point affected by this high resistance?

**(d)** Is the balance point affected by the internal resistance of the driver cell?

**(e) **Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?

**(f ) **Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

**(a) **Constant emf of the given standard cell, *E*_{1} = 1.02 V

Balance point on the wire, *l*_{1 }= 67.3 cm

A cell of unknown emf, *ε*,replaced the standard cell. Therefore, new balance point on the wire, *l* = 82.3 cm

The relation connecting emf and balance point is,

The value of unknown emfis 1.247 V.

**(b) **The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

**(c) **The balance point is not affected by the presence of high resistance.

**(d) **The point is not affected by the internal resistance of the driver cell.

**(e) **The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

**(f) **The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor *R* = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance *X *is 68.5 cm. Determine the value of *X*. What might you do if you failed to find a balance point with the given cell of emf *ε*?

Resistance of the standard resistor, *R* = 10.0 Ω

Balance point for this resistance, *l*_{1} = 58.3 cm

Current in the potentiometer wire = *i*

Hence, potential drop across *R*, *E*_{1} = *iR*

Resistance of the unknown resistor = *X*

Balance point for this resistor, *l*_{2} = 68.5 cm

Hence, potential drop across *X*, *E*_{2} = *iX*

The relation connecting emf and balance point is,

Therefore, the value of the unknown resistance, *X*, is 11.75 Ω.

If we fail to find a balance point with the given cell of emf, *ε*, then the potential drop across *R* and *X* must be reduced by putting a resistance in series with it. Only if the potential drop across *R* or *X* is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.

Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Internal resistance of the cell = *r*

Balance point of the cell in open circuit, *l*_{1} = 76.3 cm

An external resistance (*R*) is connected to the circuit with *R* = 9.5 Ω

New balance point of the circuit, *l*_{2 }= 64.8 cm

Current flowing through the circuit = *I*

The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω.

**Conclusion**: If you liked NCERT Class 12 Physics Solutions Chapter 3 – Current Electricity, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 3 – Current Electricity appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 2 – Electrostatic Potential And Capacitance appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated Class 12 Physics solutions of Chapter 1 i.e Electric Charges and Fields. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance:**

Section Name | Topic Name |

2 | Electrostatic Potential and Capacitance |

2.1 | Introduction |

2.2 | Electrostatic Potential |

2.3 | Potential due to a Point Charge |

2.4 | Potential due to an Electric Dipole |

2.5 | Potential due to a System of Charges |

2.6 | Equipotential Surfaces |

2.7 | Potential Energy of a System of Charges |

2.8 | Potential Energy in an External Field |

2.9 | Electrostatics of Conductors |

2.10 | Dielectrics and Polarisation |

2.11 | Capacitors and Capacitance |

2.12 | The Parallel Plate Capacitor |

2.13 | Effect of Dielectric on Capacitance |

2.14 | Combination of Capacitors |

2.15 | Energy Stored in a Capacitor |

2.16 | Van de Graaff Generator |

Two charges 5 × 10^{−8} C and −3 × 10^{−8} C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

There are two charges,

Distance between the two charges, *d* = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

*r* = Distance of point P from charge *q*_{1}

Let the electric potential (*V*) at point P be zero.

Potential at point P is the sum of potentials caused by charges *q*_{1} and *q*_{2} respectively.

Where,

= Permittivity of free space

For *V* = 0, equation (i) reduces to

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance *s *from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

For *V* = 0, equation (ii) reduces to

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

The given figure shows six equal amount of charges, *q*, at the vertices of a regular hexagon.

Where,

Charge, *q* = 5 µC = 5 × 10^{−6} C

Side of the hexagon, *l* = AB = BC = CD = DE = EF = FA = 10 cm

Distance of each vertex from centre O, *d* = 10 cm

Electric potential at point O,

Where,

= Permittivity of free space

Therefore, the potential at the centre of the hexagon is 2.7 × 10^{6} V.

Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.

**(a)** Identify an equipotential surface of the system.

**(b)** What is the direction of the electric field at every point on this surface?

**(a) **The situation is represented in the given figure.

An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.

**(b) **The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^{−7}C distributed uniformly on its surface. What is the electric field

**(a) **Inside the sphere

**(b) **Just outside the sphere

**(c) **At a point 18 cm from the centre of the sphere?

**(a) **Radius of the spherical conductor, *r* = 12 cm = 0.12 m

Charge is uniformly distributed over the conductor, *q *= 1.6 × 10^{−7} C

Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.

**(b) **Electric field *E* just outside the conductor is given by the relation,

Where,

= Permittivity of free space

Therefore, the electric field just outside the sphere is .

**(c) **Electric field at a point 18 m from the centre of the sphere = *E*_{1}

Distance of the point from the centre, *d* = 18 cm = 0.18 m

Therefore, the electric field at a point 18 cm from the centre of the sphere is

.

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^{−12} F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance between the parallel plates was *d *and it was filled with air. Dielectric constant of air, *k* = 1

Capacitance, *C*, is given by the formula,

Where,

*A* = Area of each plate

= Permittivity of free space

If distance between the plates is reduced to half, then new distance, *d*^{’} =

Dielectric constant of the substance filled in between the plates, = 6

Hence, capacitance of the capacitor becomes

Taking ratios of equations (i) and (ii), we obtain

Therefore, the capacitance between the plates is 96 pF.

Three capacitors each of capacitance 9 pF are connected in series.

**(a)** What is the total capacitance of the combination?

**(b)** What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

**(a) **Capacitance of each of the three capacitors, *C* = 9 pF

Equivalent capacitance (*C*^{’}) of the combination of the capacitors is given by the relation,

1C’=1C+1C+1C⇒1C’=19+19+19=13⇒C’=3 pFTherefore, total capacitance of the combination is

3 pF.

**(b) **Supply voltage, *V* = 120 V

Potential difference (*V*‘) across each capacitor is equal to one-third of the supply voltage.

Therefore, the potential difference across each capacitor is 40 V.

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

**(a)** What is the total capacitance of the combination?

**(b)** Determine the charge on each capacitor if the combination is connected to a 100 V supply.

**(a) **Capacitances of the given capacitors are

For the parallel combination of the capacitors, equivalent capacitoris given by the algebraic sum,

Therefore, total capacitance of the combination is 9 pF.

**(b) **Supply voltage, *V* = 100 V

The voltage through all the three capacitors is same = *V* = 100 V

Charge on a capacitor of capacitance *C* and potential difference *V* is given by the relation,

*q* = *VC* … (i)

For C = 2 pF,

For C = 3 pF,

For C = 4 pF,

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^{−3} m^{2} and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Area of each plate of the parallel plate capacitor, *A* = 6 × 10^{−3} m^{2}

Distance between the plates, *d* = 3 mm = 3 × 10^{−3} m

Supply voltage, *V* = 100 V

Capacitance *C* of a parallel plate capacitor is given by,

Where,

= Permittivity of free space

= 8.854 × 10^{−12} N^{−1} m^{−2} C^{−2}

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 × 10^{−9} C.

Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

**(a) **While the voltage supply remained connected.

**(b) **After the supply was disconnected.

**(a)** Dielectric constant of the mica sheet, *k* = 6

Initial capacitance, *C* = 1.771 × 10^{−11} F

Supply voltage, *V *= 100 V

Potential across the plates remains 100 V.

**(b)** Dielectric constant,* k* = 6

Initial capacitance, *C* = 1.771 × 10^{−11} F

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge = 1.771 × 10^{−9} C

Potential across the plates is given by,

A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Capacitor of the capacitance, *C* = 12 pF = 12 × 10^{−12} F

Potential difference, *V *= 50 V

Electrostatic energy stored in the capacitor is given by the relation,

Therefore, the electrostatic energy stored in the capacitor is

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Capacitance of the capacitor, *C* = 600 pF

Potential difference, *V* = 200 V

Electrostatic energy stored in the capacitor is given by,

If supply is disconnected from the capacitor and another capacitor of capacitance *C* = 600 pF is connected to it, then equivalent capacitance (*C*^{’}) of the combination is given by,

New electrostatic energy can be calculated as

Therefore, the electrostatic energy lost in the process is.

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10^{−9} C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Charge located at the origin, *q* = 8 mC= 8 × 10^{−3 }C

Magnitude of a small charge, which is taken from a point P to point R to point Q, *q*_{1} = − 2 × 10^{−9 }C

All the points are represented in the given figure.

Point P is at a distance, *d*_{1} = 3 cm, from the origin along *z*-axis.

Point Q is at a distance, *d*_{2} = 4 cm, from the origin along *y*-axis.

Potential at point P,

Potential at point Q,

Work done (*W*) by the electrostatic force is independent of the path.

Therefore, work done during the process is 1.27 J.

A cube of side *b *has a charge *q *at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Length of the side of a cube = *b*

Charge at each of its vertices = *q*

A cube of side *b *is shown in the following figure.

*d* = Diagonal of one of the six faces of the cube

*l* = Length of the diagonal of the cube

The electric potential (*V*) at the centre of the cube is due to the presence of eight charges at the vertices.

Therefore, the potential at the centre of the cube is .

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.

Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

**(a)** at the mid-point of the line joining the two charges, and

**(b)** at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, *q*_{1} = 1.5 μC

Magnitude of charge located at B, *q*_{2} = 2.5 μC

Distance between the two charges, *d* = 30 cm = 0.3 m

**(a) **Let *V*_{1} and *E*_{1} are the electric potential and electric field respectively at O.

*V*_{1} = Potential due to charge at A + Potential due to charge at B

Where,

∈_{0} = Permittivity of free space

*E*_{1} = Electric field due to *q*_{2} − Electric field due to *q*_{1}

Therefore, the potential at mid-point is 2.4 × 10^{5} V and the electric field at mid-point is 4× 10^{5} V m^{−1}. The field is directed from the larger charge to the smaller charge.

**(b) **Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.

*V*_{2} and *E*_{2 }are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

*V*_{2}= Electric potential due to A + Electric Potential due to B

Electric field due to *q* at Z,

Electric field due to *q*_{2} at *Z*,

The resultant field intensity at Z,

Where, 2*θ*is the angle, ∠*AZ *B

From the figure, we obtain

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 10^{5} V and electric field is 6.6 ×10^{5} V m^{−1}.

A spherical conducting shell of inner radius *r*1 and outer radius *r*2 has a charge *Q*.

**(a) **A charge *q *is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

**(b) **Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

**(a)** Charge placed at the centre of a shell is +*q*. Hence, a charge of magnitude −*q* will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is −*q*.

Surface charge density at the inner surface of the shell is given by the relation,

A charge of +*q* is induced on the outer surface of the shell. A charge of magnitude *Q* is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is *Q* + *q*. Surface charge density at the outer surface of the shell,

**(b) **Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

**(a)** Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

Where

is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ

**(b)** Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

**(a) **Electric field on one side of a charged body is *E*_{1} and electric field on the other side of the same body is *E*_{2}. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

Where,

= Unit vector normal to the surface at a point

σ = Surface charge density at that point

Electric field due to the other surface of the charged body,

Electric field at any point due to the two surfaces,

Since inside a closed conductor, = 0,

∴

Therefore, the electric field just outside the conductor is .

**(b) **When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Charge density of the long charged cylinder of length *L* and radius *r* is *λ*.

Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is *R*.

Let *E* be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss’s theorem as,

Where, *d *= Distance of a point from the common axis of the cylinders

Let *q* be the total charge on the cylinder.

It can be written as

Where,

*q* = Charge on the inner sphere of the outer cylinder

∈_{0} = Permittivity of free space

Therefore, the electric field in the space between the two cylinders is.

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

**(a)** Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

**(b)** What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

**(c)** What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

The distance between electron-proton of a hydrogen atom,

Charge on an electron, *q*_{1} = −1.6 ×10^{−19} C

Charge on a proton, *q*_{2} = +1.6 ×10^{−19} C

**(a)** Potential at infinity is zero.

Potential energy of the system,= Potential energy at infinity − Potential energy at distance *d*

where,

∈_{0} is the permittivity of free space

14πε0=9×109 Nm2C-2∴ Potential energy=0-9×109×1.6×10-1920.53×10-10=-43.47×10-19 J∵1.6×10-19 J=1 eV∴Potential energy=-43.7×10-19=-43.7×10-191.6×10-19=-27.2 eV

Therefore, the potential energy of the system is −27.2 eV.

**(b)** Kinetic energy is half of the magnitude of potential energy.

Total energy = 13.6 − 27.2 = 13.6 eV

Therefore, the minimum work required to free the electron is 13.6 eV.

**(c)** When zero of potential energy is taken,

∴Potential energy of the system = Potential energy at *d*_{1} − Potential energy at *d*

If one of the two electrons of a H_{2 }molecule is removed, we get a hydrogen molecular ion. In the ground state of an, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

The system of two protons and one electron is represented in the given figure.

Charge on proton 1, *q*_{1} = 1.6 ×10^{−19 }C

Charge on proton 2, *q*_{2} = 1.6 ×10^{−19 }C

Charge on electron, *q*_{3} = −1.6 ×10^{−19 }C

Distance between protons 1 and 2, *d*_{1} = 1.5 ×10^{−10} m

Distance between proton 1 and electron, *d*_{2} = 1 ×10^{−10} m

Distance between proton 2 and electron, *d*_{3} = 1 × 10^{−10} m

The potential energy at infinity is zero.

Potential energy of the system,

Therefore, the potential energy of the system is −19.2 eV.

Two charged conducting spheres of radii *a* and *b* are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Let *a* be the radius of a sphere A, *Q*_{A} be the charge on the sphere, and *C*_{A} be the capacitance of the sphere. Let *b* be the radius of a sphere B, *Q*_{B} be the charge on the sphere, and *C*_{B} be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (*V*) will become equal.

Let *E*_{A}be the electric field of sphere A and *E*_{B} be the electric field of sphere B. Therefore, their ratio,

Putting the value of (2) in (1), we obtain

Therefore, the ratio of electric fields at the surface is.

A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.

Two charges *−q *and *+q *are located at points (0, 0, − *a*) and (0, 0, *a*), respectively.

**(a)** What is the electrostatic potential at the points?

**(b)** Obtain the dependence of potential on the distance *r *of a point from the origin when *r*/*a *>> 1.

**(c)** How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the *x*-axis? Does the answer change if the path of the test charge between the same points is not along the *x*-axis?

**(a) **Zero at both the points

Charge − *q* is located at (0, 0, − *a*) and charge + *q* is located at (0, 0, *a*). Hence, they form a dipole. Point (0, 0, *z*) is on the axis of this dipole and point (*x*, *y*, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (*x*, *y*, 0) is zero. Electrostatic potential at point (0, 0, *z*) is given by,

Where,

= Permittivity of free space

*p* = Dipole moment of the system of two charges = 2*qa*

**(b)** Distance *r* is much greater than half of the distance between the two charges. Hence, the potential (*V*) at a distance *r* is inversely proportional to square of the distance i.e.,

**(c) **Zero

The answer does not change if the path of the test is not along the *x*-axis.

A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the *x*-axis. Electrostatic potential (*V*_{1}) at point (5, 0, 0) is given by,

V1=-q4π∈01(5-0)2+(-a)2+ q4π∈01(5-0)2+(a)2 =-q4π∈025+a2+q4π∈025+a2 =0Electrostatic potential, *V*_{2}, at point (− 7, 0, 0) is given by,

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the *x*-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

Figure 2.34 shows a charge array known as an *electric quadrupole*. For a point on the axis of the quadrupole, obtain the dependence of potential on *r *for *r*/*a *>> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.

A point is located at P, which is *r* distance away from point Y.

The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge +*q *placed at point X

Charge −2*q* placed at point Y

Charge +*q* placed at point Z

XY = YZ = *a*

YP =* r*

PX = *r* + *a*

PZ = *r* − *a*

Electrostatic potential caused by the system of three charges at point P is given by,

Since,

is taken as negligible.

It can be inferred that potential,

However, it is known that for a dipole,

And, for a monopole,

An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Total required capacitance, *C* = 2 µF

Potential difference, *V* = 1 kV = 1000 V

Capacitance of each capacitor, *C*_{1} = 1µF

Each capacitor can withstand a potential difference, *V*_{1} = 400 V

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as

Hence, there are three capacitors in each row.

Capacitance of each row

Let there are *n* rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Capacitance of a parallel capacitor, *V* = 2 F

Distance between the two plates, *d* = 0.5 cm = 0.5 × 10^{−2} m

Capacitance of a parallel plate capacitor is given by the relation,

Where,

= Permittivity of free space = 8.85 × 10^{−12} C^{2} N^{−1} m^{−2}

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.

Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

Capacitance of capacitor *C*_{1 }is 100 pF.

Capacitance of capacitor *C*_{2 }is 200 pF.

Capacitance of capacitor *C*_{3 }is 200 pF.

Capacitance of capacitor *C*_{4 }is 100 pF.

Supply potential, *V* = 300 V

Capacitors *C*_{2} and *C*_{3} are connected in series. Let their equivalent capacitance be

Capacitors *C*_{1} and *C’ *are in parallel. Let their equivalent capacitance be

are connected in series. Let their equivalent capacitance be *C*.

Hence, the equivalent capacitance of the circuit is

Potential difference across =

Potential difference across *C*_{4} =* V*_{4}

Charge on

*Q*_{4}= *CV*

Hence, potential difference, *V*_{1}, across *C*_{1} is 100 V.

Charge on *C*_{1} is given by,

*C*_{2} and *C*_{3} having same capacitances have a potential difference of 100 V together. Since *C*_{2} and *C*_{3} are in series, the potential difference across *C*_{2 }and *C*_{3 }is given by,

*V*_{2}* = V*_{3}* = *50 V

Therefore, charge on *C*_{2} is given by,

And charge on *C*_{3 }is given by,

Hence, the equivalent capacitance of the given circuit is

The plates of a parallel plate capacitor have an area of 90 cm^{2} each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

**(a)** How much electrostatic energy is stored by the capacitor?

**(b)** View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume *u*. Hence arrive at a relation between *u *and the magnitude of electric field *E *between the plates.

Area of the plates of a parallel plate capacitor, *A* = 90 cm^{2 }= 90 × 10^{−4 }m^{2}

Distance between the plates, *d* = 2.5 mm = 2.5 × 10^{−3 }m

Potential difference across the plates, *V* = 400 V

**(a) **Capacitance of the capacitor is given by the relation,

Electrostatic energy stored in the capacitor is given by the relation,

Where,

= Permittivity of free space = 8.85 × 10^{−12} C^{2} N^{−1} m^{−2}

Hence, the electrostatic energy stored by the capacitor is

**(b) **Volume of the given capacitor,

Energy stored in the capacitor per unit volume is given by,

Where,

= Electric intensity = *E*

A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Capacitance of a charged capacitor,

Supply voltage, *V*_{1} = 200 V

Electrostatic energy stored in *C*_{1 }is given by,

Capacitance of an uncharged capacitor,

When *C*_{2} is connected to the circuit, the potential acquired by it is *V*_{2}.

According to the conservation of charge, initial charge on capacitor *C*_{1} is equal to the final charge on capacitors, *C*_{1} and *C*_{2}.

Electrostatic energy for the combination of two capacitors is given by,

Hence, amount of electrostatic energy lost by capacitor *C*_{1}

= *E*_{1} − *E*_{2}

= 0.08 − 0.0533 = 0.0267

= 2.67 × 10^{−2} J

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) *QE*, where *Q *is the charge on the capacitor, and *E *is the magnitude of electric field between the plates. Explain the origin of the factor ½.

Let *F* be the force applied to separate the plates of a parallel plate capacitor by a distance of Δ*x*. Hence, work done by the force to do so = *F**Δ**x*

As a result, the potential energy of the capacitor increases by an amount given as *uA**Δ**x*.

Where,

*u* = Energy density

*A* = Area of each plate

*d* = Distance between the plates

*V* = Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

Electric intensity is given by,

However, capacitance,

Charge on the capacitor is given by,

*Q* = *CV*

The physical origin of the factor, , in the force formula lies in the fact that just outside the conductor, field is *E* and inside it is zero. Hence, it is the average value, , of the field that contributes to the force.

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show

that the capacitance of a spherical capacitor is given by

where *r*_{1} and *r*_{2} are the radii of outer and inner spheres, respectively.

Radius of the outer shell = *r*_{1}

Radius of the inner shell = *r*_{2}

The inner surface of the outer shell has charge +*Q*.

The outer surface of the inner shell has induced charge −*Q*.

Potential difference between the two shells is given by,

Where,

= Permittivity of free space

Hence, proved.

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

**(a)** Determine the capacitance of the capacitor.

**(b)** What is the potential of the inner sphere?

**(c)** Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Radius of the inner sphere, = 12 cm = 0.12 m

Radius of the outer sphere, = 13 cm = 0.13 m

Charge on the inner sphere,

Dielectric constant of a liquid,

**(a) **

Where,

= Permittivity of free space =

Hence, the capacitance of the capacitor is approximately .

**(b) **Potential of the inner sphere is given by,

Hence, the potential of the inner sphere is .

**(c) **Radius of an isolated sphere, *r *= 12 × 10^{−2} m

Capacitance of the sphere is given by the relation,

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

Answer carefully:

**(a) **Two large conducting spheres carrying charges *Q*_{1} and *Q*_{2} are brought close to each other. Is the magnitude of electrostatic force between them exactly given by *Q*_{1}* Q*_{2}/4π*r *^{2}, where *r *is the distance between their centres?

**(b)** If Coulomb’s law involved 1/*r*^{3} dependence (instead of 1/*r*^{2}), would Gauss’s law be still true?

**(c)** A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

**(d)** What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

**(e)** We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

**(f)** What meaning would you give to the capacitance of a single conductor?

**(g)** Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

**(a) **The force between two conducting spheres is not exactly given by the expression, *Q*_{1}* Q*_{2}/4π*r *^{2}, because there is a non-uniform charge distribution on the spheres.

**(b) **Gauss’s law will not be true, if Coulomb’s law involved 1/*r*^{3 }dependence, instead of1/*r*^{2}, on *r*.

**(c) **Yes,

If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

**(d) **Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

**(e) **No

Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

**(f) **The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

**(g) **Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Length of a co-axial cylinder,* l* = 15 cm = 0.15 m

Radius of outer cylinder, *r*_{1} = 1.5 cm = 0.015 m

Radius of inner cylinder, *r*_{2} = 1.4 cm = 0.014 m

Charge on the inner cylinder, *q* = 3.5 µC = 3.5 × 10^{−6} C

Where,

= Permittivity of free space =

Potential difference of the inner cylinder is given by,

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10^{7} Vm^{−1}. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Potential rating of a parallel plate capacitor, *V* = 1 kV = 1000 V

Dielectric constant of a material, =3

Dielectric strength = 10^{7} V/m

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field intensity, *E* = 10% of 10^{7 }= 10^{6} V/m

Capacitance of the parallel plate capacitor, *C* = 50 pF = 50 × 10^{−12} F

Distance between the plates is given by,

Where,

*A* = Area of each plate

= Permittivity of free space =

Hence, the area of each plate is about 19 cm^{2}.

Describe schematically the equipotential surfaces corresponding to

**(a)** a constant electric field in the *z*-direction,

**(b)** a field that uniformly increases in magnitude but remains in a constant (say, *z*) direction,

**(c)** a single positive charge at the origin, and

**(d)** a uniform grid consisting of long equally spaced parallel charged wires in a plane.

**(a) **Equidistant planes parallel to the *x*–*y* plane are the equipotential surfaces.

**(b) **Planes parallel to the *x-y *plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

**(c) **Concentric spheres centered at the origin are equipotential surfaces.

**(d) **A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

In a Van de Graaff type generator a spherical metal shell is to be a 15 × 10^{6} V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10^{7} Vm^{−1}. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Potential difference, *V* = 15 × 10^{6} V

Dielectric strength of the surrounding gas = 5 × 10^{7} V/m

Electric field intensity, *E* = Dielectric strength = 5 × 10^{7} V/m

Minimum radius of the spherical shell required for the purpose is given by,

Hence, the minimum radius of the spherical shell required is 30 cm.

A small sphere of radius *r*_{1} and charge *q*_{1} is enclosed by a spherical shell of radius *r*_{2} and charge *q*_{2}. Show that if *q*_{1} is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge *q*_{2} on the shell is.

According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge *q*_{1} on a small sphere. Hence, the potential difference, *V*, between the sphere and the shell is independent of charge *q*_{2}.For positive charge *q*_{1}, potential difference *V *is always positive.

Answer the following:

**(a)** The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^{−1}. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

**(b)** A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m^{2}. Will he get an electric shock if he touches the metal sheet next morning?

**(c)** The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

**(d)** What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm^{−1} at its surface in the downward direction, corresponding to a surface charge density = −10^{−9} C m^{−2}. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

**(a) **We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

**(b) **Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

**(c) **The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

**(d) **During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.

**Conclusion**: If you liked NCERT Class 12 Physics Solutions Chapter 2 – Electrostatic Potential and Capacitance, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 2 – Electrostatic Potential And Capacitance appeared first on Babaji Academy.

]]>The post NCERT Class 12 Physics Solutions Chapter 1 – Electric Charges and Fields appeared first on Babaji Academy.

]]>We had already updated solutions of Maths for class 12 All chapters, you can check it from here. If you have not checked it yet, then check it now. You can download pdf of all these chapters in your mobile/laptop.

Topics and Subtopics in **NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges And Fields:**

Section Name | Topic Name |

1 | Electric Charges And Fields |

1.1 | Introduction |

1.2 | Electric Charge |

1.3 | Conductors and Insulators |

1.4 | Charging by Induction |

1.5 | Basic Properties of Electric Charge |

1.6 | Coulomb’s Law |

1.7 | Forces between Multiple Charges |

1.8 | Electric Field |

1.9 | Electric Field Lines |

1.10 | Electric Flux |

1.11 | Electric Dipole |

1.12 | Dipole in a Uniform External Field |

1.13 | Continuous Charge Distribution |

1.14 | Gauss’s Law |

1.15 | Applications of Gauss’s Law |

What is the force between two small charged spheres having charges of 2 × 10^{−7} C and 3 × 10^{−7} C placed 30 cm apart in air?

Repulsive force of magnitude 6 × 10^{−3} N

Charge on the first sphere, *q*_{1} = 2 × 10^{−7} C

Charge on the second sphere, *q*_{2} = 3 × 10^{−7} C

Distance between the spheres, *r *= 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

Where, ∈_{0} = Permittivity of free space

Hence, force between the two small charged spheres is 6 × 10^{−3} N. The charges are of same nature. Hence, force between them will be repulsive.

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

**(a) **Electrostatic force on the first sphere, *F* = 0.2 N

Charge on this sphere, *q*_{1} = 0.4 μC = 0.4 × 10^{−6 }C

Charge on the second sphere, *q*_{2} = − 0.8 μC = − 0.8 × 10^{−6} C

Electrostatic force between the spheres is given by the relation,

Where, ∈_{0} = Permittivity of free space

r2 = q1q24π∈0F = 9×109×0.4×10-6×0.8×10-60.2 = 144×10-4r = 144×10-4 = 0.12 m

The distance between the two spheres is 0.12 m.

**(b) **Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.

Check that the ratio *ke*^{2}/*G m*_{e}*m*_{p}is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

The given ratio is.

Where,

G = Gravitational constant

Its unit is N m^{2 }kg^{−2}.

*m*_{e} and *m*_{p} = Masses of electron and proton.

Their unit is kg.

*e* = Electric charge.

Its unit is C.

∈_{0} = Permittivity of free space

Its unit is N m^{2 }C^{−2}.

Hence, the given ratio is dimensionless.

*e *= 1.6 × 10^{−19 }C

G = 6.67 × 10^{−11} N m^{2 }kg^{-2}

*m*_{e}= 9.1 × 10^{−31 }kg

*m*_{p} = 1.66 × 10^{−27 }kg

Hence, the numerical value of the given ratio is

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

**(a)** Explain the meaning of the statement ‘electric charge of a body is quantised’.

**(b)** Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

**(a)** Electric charge of a body is quantized. This means that only integral (1, 2, …., *n*) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

**(b) **In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Four point charges *q*_{A }= 2 μC, *q*_{B }= −5 μC, *q*_{C} = 2 μC, and *q*_{D} = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

Where,

(Sides) AB = BC = CD = AD = 10 cm

(Diagonals) AC = BD = cm

AO = OC = DO = OB = cm

A charge of amount 1μC is placed at point O.

Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.

**(a)** An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

**(b)** Explain why two field lines never cross each other at any point?

**(a)** An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

**(b)** If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Two point charges *q*_{A} = 3 μC and *q*_{B} = −3 μC are located 20 cm apart in vacuum.

**(a)** What is the electric field at the midpoint O of the line AB joining the two charges?

**(b)** If a negative test charge of magnitude 1.5 × 10^{−9} C is placed at this point, what is the force experienced by the test charge?

**(a) **The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm

∴AO = OB = 10 cm

Net electric field at point O = *E*

Electric field at point O caused by +3μC charge,

*E*_{1} = along OB

Where,

= Permittivity of free space

Magnitude of electric field at point O caused by −3μC charge,

*E*_{2} = = along OB

= 5.4 × 10^{6} N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 10^{6} N C^{−1} along OB.

**(b) **A test charge of amount 1.5 × 10^{−9} C is placed at mid-point O.

*q* = 1.5 × 10^{−9} C

Force experienced by the test charge = *F*

∴*F = qE*

= 1.5 × 10^{−9} × 5.4 × 10^{6}

= 8.1 × 10^{−3} N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 × 10^{−3} N along OA.

A system has two charges *q*_{A} = 2.5 × 10^{−7} C and *q*_{B} = −2.5 × 10^{−7} C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, *q*_{A} = 2.5 × 10^{−}^{7}C

At B, amount of charge, *q*_{B} = −2.5 × 10^{−7} C

Total charge of the system,

*q* = *q*_{A} + *q*_{B}

= 2.5 × 10^{7} C − 2.5 × 10^{−7} C

= 0

Distance between two charges at points A and B,

*d* = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

*p* = *q*_{A} × *d =* *q*_{B} × *d*

= 2.5 × 10^{−7} × 0.3

= 7.5 × 10^{−8} C m along positive *z*-axis

Therefore, the electric dipole moment of the system is 7.5 × 10^{−8} C m along positive *z*−axis.

An electric dipole with dipole moment 4 × 10^{−9} C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10^{4} N C^{−1}. Calculate the magnitude of the torque acting on the dipole.

Electric dipole moment, *p* = 4 × 10^{−9} C m

Angle made by *p* with a uniform electric field, *θ* = 30°

Electric field, *E* = 5 × 10^{4} N C^{−1}

Torque acting on the dipole is given by the relation,

τ = *pE* sin*θ*

Therefore, the magnitude of the torque acting on the dipole is 10^{−4} N m.

A polythene piece rubbed with wool is found to have a negative charge of 3 × 10^{−7} C.

**(a)** Estimate the number of electrons transferred (from which to which?)

**(b)** Is there a transfer of mass from wool to polythene?

**(a) **When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.

Amount of charge on the polythene piece, *q* = −3 × 10^{−7} C

Amount of charge on an electron, *e* = −1.6 × 10^{−19} C

Number of electrons transferred from wool to polythene = *n*

*n *can be calculated using the relation,

*q* = *ne*

= 1.87 × 10^{12}

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 10^{12}.

**(b) **Yes.

There is a transfer of mass taking place. This is because an electron has mass,

*m*_{e} = 9.1 × 10^{−3} kg

Total mass transferred to polythene from wool,

*m* = *m*_{e}* × n*

= 9.1 × 10^{−31} × 1.85 × 10^{12}

= 1.706 × 10^{−18} kg

Hence, a negligible amount of mass is transferred from wool to polythene.

**(a)** Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^{−7} C? The radii of A and B are negligible compared to the distance of separation.

**(b)** What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

**(a) **Charge on sphere A, *q*_{A} = Charge on sphere B, *q*_{B} = 6.5 × 10^{−7} C

Distance between the spheres, *r* = 50 cm = 0.5 m

Force of repulsion between the two spheres,

Where,

∈_{0} = Free space permittivity

= 9 × 10^{9} N m^{2} C^{−2}

∴

= 1.52 × 10^{−2} N

Therefore, the force between the two spheres is 1.52 × 10^{−2} N.

**(b) **After doubling the charge, charge on sphere A, *q*_{A} = Charge on sphere B, *q*_{B} = 2 × 6.5 × 10^{−7} C = 1.3 × 10^{−6} C

The distance between the spheres is halved.

∴

Force of repulsion between the two spheres,

= 16 × 1.52 × 10^{−2}

= 0.243 N

Therefore, the force between the two spheres is 0.243 N.

Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Distance between the spheres, A and B, *r* = 0.5 m

Initially, the charge on each sphere, *q* = 6.5 × 10^{−7} C

When sphere A is touched with an uncharged sphere C, amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is.

When sphere C with charge is brought in contact with sphere B with charge *q*, total charges on the system will divide into two equal halves given as,

Each sphere will share each half. Hence, charge on each of the spheres, C and B, is.

Force of repulsion between sphere A having charge and sphere B having charge =

Therefore, the force of attraction between the two spheres is 5.703 × 10^{−3} N.

Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Consider a uniform electric field **E **= 3 × 10^{3} îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the *yz *plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the *x*-axis?

**(a) **Electric field intensity, = 3 × 10^{3 }î N/C

Magnitude of electric field intensity, = 3 × 10^{3} N/C

Side of the square, *s* = 10 cm = 0.1 m

Area of the square, *A* = s^{2} = 0.01 m^{2}

The plane of the square is parallel to the *y-z* plane. Hence, angle between the unit vector normal to the plane and electric field, *θ* = 0°

Flux (*Φ*) through the plane is given by the relation,

Φ =

= 3 × 10^{3} × 0.01 × cos0°

= 30 N m^{2}/C

**(b) **Plane makes an angle of 60° with the *x*-axis. Hence, *θ* = 60°

Flux, *Φ* =

= 3 × 10^{3} × 0.01 × cos60°

= 15 N m^{2}/C

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10^{3} N m^{2}/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

**(a) **Net outward flux through the surface of the box, *Φ* = 8.0 × 10^{3} N m^{2}/C

For a body containing net charge *q*, flux is given by the relation,

∈_{0} = Permittivity of free space

= 8.854 × 10^{−12} N^{−1}C^{2 }m^{−2}

*q* = ∈_{0}*Φ*

= 8.854 × 10^{−12} × 8.0 × 10^{3}

= 7.08 × 10^{−8}

= 0.07 μC

Therefore, the net charge inside the box is 0.07 μC.

**(b) **No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (*Hint: *Think of the square as one face of a cube with edge 10 cm.)

The square can be considered as one face of a cube of edge 10 cm with a centre where charge *q* is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

Hence, electric flux through one face of the cube i.e., through the square,

Where,

∈_{0} = Permittivity of free space

= 8.854 × 10^{−12} N^{−1}C^{2 }m^{−2}

*q* = 10 μC = 10 × 10^{−6} C

∴

= 1.88 × 10^{5} N m^{2} C^{−1}

Therefore, electric flux through the square is 1.88 × 10^{5} N m^{2} C^{−1}.

A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Net electric flux (*Φ*_{Net}) through the cubic surface is given by,

Where,

∈_{0} = Permittivity of free space

= 8.854 × 10^{−12} N^{−1}C^{2 }m^{−2}

*q* = Net charge contained inside the cube = 2.0 μC = 2 × 10^{−6} C

∴

= 2.26 × 10^{5} N m^{2} C^{−1}

The net electric flux through the surface is 2.26 ×10^{5} N m^{2}C^{−1}.

A point charge causes an electric flux of −1.0 × 10^{3} Nm^{2}/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

**(a) **Electric flux, *Φ* = −1.0 × 10^{3} N m^{2}/C

Radius of the Gaussian surface,

*r* = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −10^{3} N m^{2}/C.

**(b) **Electric flux is given by the relation,

Where,

*q* = Net charge enclosed by the spherical surface

∈_{0} = Permittivity of free space = 8.854 × 10^{−12} N^{−1}C^{2 }m^{−2}

∴

= −1.0 × 10^{3} × 8.854 × 10^{−12}

= −8.854 × 10^{−9} C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10^{3 }N/C and points radially inward, what is the net charge on the sphere?

Electric field intensity (*E*) at a distance (*d*) from the centre of a sphere containing net charge *q* is given by the relation,

Where,

*q* = Net charge = 1.5 × 10^{3} N/C

*d *= Distance from the centre = 20 cm = 0.2 m

∈_{0} = Permittivity of free space

And, = 9 × 10^{9} N m^{2} C^{−2}

∴

= 6.67 × 10^{9} C

= 6.67 nC

Therefore, the net charge on the sphere is 6.67 nC.

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m^{2}. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

**(a) **Diameter of the sphere, *d* = 2.4 m

Radius of the sphere, *r* = 1.2 m

Surface charge density, = 80.0 μC/m^{2} = 80 × 10^{−6} C/m^{2}

Total charge on the surface of the sphere,

*Q* = Charge density × Surface area

=

= 80 × 10^{−6} × 4 × 3.14 × (1.2)^{2}

= 1.447 × 10^{−3} C

Therefore, the charge on the sphere is 1.447 × 10^{−3} C.

**(b) **Total electric flux () leaving out the surface of a sphere containing net charge *Q *is given by the relation,

Where,

∈_{0} = Permittivity of free space

= 8.854 × 10^{−12} N^{−1}C^{2 }m^{−2}

*Q *= 1.447 × 10^{−3} C

= 1.63 × 10^{8} N C^{−1} m^{2}

Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10^{8} N C^{−1} m^{2}.

An infinite line charge produces a field of 9 × 10^{4} N/C at a distance of 2 cm. Calculate the linear charge density.

Electric field produced by the infinite line charges at a distance *d* having linear charge density *λ* is given by the relation,

Where,

*d* = 2 cm = 0.02 m

*E* = 9 × 10^{4} N/C

∈_{0} = Permittivity of free space

= 9 × 10^{9} N m^{2} C^{−2}

= 10 μC/m

Therefore, the linear charge density is 10 μC/m.

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^{−22 }C/m^{2}. What is **E**: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

The situation is represented in the following figure.

A and B are two parallel plates close to each other. Outer region of plate A is labelled as **I**, outer region of plate B is labelled as **III**, and the region between the plates, A and B, is labelled as **II.**

Charge density of plate A, *σ* = 17.0 × 10^{−22} C/m^{2}

Charge density of plate B, *σ* = −17.0 × 10^{−22} C/m^{2}

In the regions, **I** and **III**, electric field *E* is zero. This is because charge is not enclosed by the respective plates.

Electric field *E* in region **II** is given by the relation,

Where,

∈_{0} = Permittivity of free space = 8.854 × 10^{−12} N^{−1}C^{2 }m^{−2}

∴

= 1.92 × 10^{−10} N/C

Therefore, electric field between the plates is 1.92 × 10^{−10} N/C.

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10^{4} N C^{−1} in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^{−3}. Estimate the radius of the drop. (*g *= 9.81 m s^{−2}; *e *= 1.60 × 10^{−19} C).

Excess electrons on an oil drop, *n* = 12

Electric field intensity, *E* = 2.55 × 10^{4} N C^{−1}

Density of oil, *ρ* = 1.26 gm/cm^{3} = 1.26 × 10^{3} kg/m^{3}

Acceleration due to gravity, g = 9.81 m s^{−2}

Charge on an electron, *e* = 1.6 × 10^{−19} C

Radius of the oil drop = *r*

Force (*F*) due to electric field *E* is equal to the weight of the oil drop (*W*)

*F = W*

*Eq *= *m*g

*Ene*

Where,

*q* = Net charge on the oil drop = *ne*

*m *= Mass of the oil drop

= Volume of the oil drop × Density of oil

= 9.82 × 10^{−4} mm

Therefore, the radius of the oil drop is 9.82 × 10^{−4} mm.

Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

**(a)**

**(b)**

**(c)**

**(d)**

**(e)**

**(a) **The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

**(b) **The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.

**(c) **The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.

**(d) **The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

**(e) **The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive *z*-direction, at the rate of 10^{5} NC^{−1} per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^{−7} Cm in the negative *z*-direction?

Dipole moment of the system, *p* = *q × dl *= −10^{−7} C m

Rate of increase of electric field per unit length,

Force (*F*) experienced by the system is given by the relation,

*F* = *qE*

= −10^{−7} × 10^{−5}

= −10^{−2} N

The force is −10^{−2} N in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180°.

Torque (*τ*) is given by the relation,

τ = *pE* sin180°

= 0

Therefore, the torque experienced by the system is zero.

(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge *Q*. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge *q *is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is *Q *+ *q *[Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

**(a) **Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity *E* inside the charged conductor is zero.

Let *q* is the charge inside the conductor and is the permittivity of free space.

According to Gauss’s law,

Flux,

Here, *E *= 0

Therefore, charge inside the conductor is zero.

The entire charge *Q* appears on the outer surface of the conductor.

**(b) **The outer surface of conductor A has a charge of amount *Q*. Another conductor B having charge +*q* is kept inside conductor A and it is insulated from A. Hence, a charge of amount −*q *will be induced in the inner surface of conductor A and +*q* is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is* Q *+ *q*.

**(c) **A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is , where is the unit vector in the outward normal direction, and is the surface charge density near the hole.

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let *E* is the electric field just outside the conductor, *q* is the electric charge, is the charge density, and is the permittivity of free space.

Charge

According to Gauss’s law,

Therefore, the electric field just outside the conductor is. This field is a superposition of field due to the cavity and the field due to the rest of the charged conductor. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.

Therefore, the field due to the rest of the conductor is.

Hence, proved.

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density *λ* without using Gauss’s law. [*Hint: *Use Coulomb’s law directly and evaluate the necessary integral.]

Take a long thin wire XY (as shown in the figure) of uniform linear charge density.

Consider a point A at a perpendicular distance *l* from the mid-point O of the wire, as shown in the following figure.

Let *E* be the electric field at point A due to the wire, XY.

Consider a small length element *dx* on the wire section with OZ = *x*

Let *q* be the charge on this piece.

Electric field due to the piece,

The electric field is resolved into two rectangular components. is the perpendicular component and is the parallel component.

When the whole wire is considered, the component is cancelled.

Only the perpendicular component affects point A.

Hence, effective electric field at point A due to the element *dx* is *dE*_{1}.

On differentiating equation (2), we obtain

dxdθ = lsec2θdx = lsec2θ dθ

From equation (2),

Putting equations (3) and (4) in equation (1), we obtain

The wire is so long that tends from to .

By integrating equation (5), we obtain the value of field *E*_{1} as,

Therefore, the electric field due to long wire is.

It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) *e*, and the ‘down’ quark (denoted by d) of charge (−1/3) *e*, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

A proton has three quarks. Let there be *n* up quarks in a proton, each having a charge of .

Charge due to *n *up quarks

Number of down quarks in a proton = 3 − *n*

Each down quark has a charge of .

Charge due to (3 −* n*) down quarks

Total charge on a proton = + *e*

Number of up quarks in a proton, *n* = 2

Number of down quarks in a proton = 3 − *n* = 3 − 2 = 1

Therefore, a proton can be represented as ‘uud’.

A neutron also has three quarks. Let there be *n* up quarks in a neutron, each having a charge of .

Charge on a neutron due to *n* up quarks

Number of down quarks is 3 − *n*,each having a charge of .

Charge on a neutron due to down quarks =

Total charge on a neutron = 0

Number of up quarks in a neutron, *n* = 1

Number of down quarks in a neutron = 3 − *n* = 2

Therefore, a neutron can be represented as ‘udd’.

**(a)** Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where **E **= 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

**(b)** Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

**(a) **Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

**(b) **Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

A particle of mass *m *and charge (−*q*) enters the region between the two charged plates initially moving along *x*-axis with speed *vx *(like particle 1 in Fig. 1.33). The length of plate is *L *and an uniform electric field *E *is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is *qEL*^{2}/ (2*m*).

*Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.*

Charge on a particle of mass *m* = − *q*

Velocity of the particle = *v*_{x}

Length of the plates =* L*

Magnitude of the uniform electric field between the plates = *E*

Mechanical force, *F* = Mass (*m*) × Acceleration (*a*)

Therefore, acceleration,

Time taken by the particle to cross the field of length *L *is given by,

*t*

In the vertical direction, initial velocity, *u* = 0

According to the third equation of motion, vertical deflection *s *of the particle can be obtained as,

Hence, vertical deflection of the particle at the far edge of the plate is

. This is similar to the motion of horizontal projectiles under gravity.

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity *v*_{x}= 2.0 × 10^{6} m s^{−1}. If *E *between the plates separated by 0.5 cm is 9.1 × 10^{2} N/C, where will the electron strike the upper plate? (| *e *| =1.6 × 10^{−19} C, *m*_{e }= 9.1 × 10^{−31} kg.)

Velocity of the particle, *v*_{x} = 2.0 × 10^{6} m/s

Separation of the two plates, *d* = 0.5 cm = 0.005 m

Electric field between the two plates, *E* = 9.1 × 10^{2} N/C

Charge on an electron, *q* = 1.6 × 10^{−19} C

Mass of an electron, *m*_{e }= 9.1 × 10^{−31} kg

Let the electron strike the upper plate at the end of plate *L*, when deflection is *s*.

Therefore,

Therefore, the electron will strike the upper plate after travelling 1.6 cm.

**Conclusion**: If you liked NCERT Class 12 Physics Solutions Chapter 1 – Electric Charges and Fields, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Physics Solutions Chapter 1 – Electric Charges and Fields appeared first on Babaji Academy.

]]>The post NCERT Class 12 Maths Solutions Chapter 13 – Probability appeared first on Babaji Academy.

]]>In our previous post, we have updated solution of Chapter 9 i.e Linear Programming. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

**The topics and sub-topics included in Chapter 13 Probability the following:**

Section Name | Topic Name |

13 | Probability |

13.1 | Introduction |

13.2 | Conditional Probability |

13.3 | Multiplication Theorem on Probability |

13.4 | Independent Events |

13.5 | Bayes’ Theorem |

13.6 | Random Variables and its Probability Distributions |

13.7 | Bernoulli Trials and Binomial Distribution |

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).

It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2

Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

It is given that P(B) = 0.5 and P(A ∩ B) = 0.32

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find

(i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4

(i) P (B|A) = 0.4

(ii)

(iii)

PA∪B = PA + PB – PA∩B⇒PA∪B=0.8 + 0.5 – 0.32 = 0.98

Evaluate P (A ∪ B), if 2P (A) = P (B) =and P(A|B) =

It is given that,

It is known that,

If P(A), P(B) =and P(A ∪ B) =, find

(i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A)

It is given that

(i)

(ii) It is known that,

(iii) It is known that,

A coin is tossed three times, where

(i) E: head on third toss, F: heads on first two tosses

(ii) E: at least two heads, F: at most two heads

(iii) E: at most two tails, F: at least one tail

If a coin is tossed three times, then the sample space S is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space has 8 elements.

(i) E = {HHH, HTH, THH, TTH}

F = {HHH, HHT}

E ∩ F = {HHH}

(ii) E = {HHH, HHT, HTH, THH}

F = {HHT, HTH, HTT, THH, THT, TTH, TTT}

E ∩ F = {HHT, HTH, THH}

Clearly,

(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}

F = {HHT, HTT, HTH, THH, THT, TTH, TTT}

Two coins are tossed once, where

(i) E: tail appears on one coin, F: one coin shows head

(ii) E: not tail appears, F: no head appears

If two coins are tossed once, then the sample space S is

S = {HH, HT, TH, TT}

(i) E = {HT, TH}

F = {HT, TH}

(ii) E = {HH}

F = {TT}

∴ E ∩ F = Φ

P (F) = 1 and P (E ∩ F) = 0

∴ P(E|F) =

A die is thrown three times,

E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses

If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216

Mother, father and son line up at random for a family picture

E: son on one end, F: father in middle

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be

S = {MFS, MSF, FMS, FSM, SMF, SFM}

⇒ E = {MFS, FMS, SMF, SFM}

F = {MFS, SFM}

∴ E ∩ F = {MFS, SFM}

A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Let the first observation be from the black die and second from the red die.

When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.

- Let

A: Obtaining a sum greater than 9

= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

B: Black die results in a 5.

= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

∴ A ∩ B = {(5, 5), (5, 6)}

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).

PA|B = PA∩BPB = 236636 = 26 = 13

(b) E: Sum of the observations is 8.

= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

F: Red die resulted in a number less than 4.

The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).

A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}

Find

(i) P (E|F) and P (F|E) (ii) P (E|G) and P (G|E)

(ii) P ((E ∪ F)|G) and P ((E ∩ G)|G)

When a fair die is rolled, the sample space S will be

S = {1, 2, 3, 4, 5, 6}

It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}

(i) E ∩ F = {3}

(ii) E ∩ G = {3, 5}

(iii) E ∪ F = {1, 2, 3, 5}

(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}

E ∩ F = {3}

(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Let *b* and *g* represent the boy and the girl child respectively. If a family has two children, the sample space will be

S = {(*b*, *b*), (*b*, *g*), (*g*, *b*), (*g,* g)}

Let A be the event that both children are girls.

(i) Let B be the event that the youngest child is a girl.

The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).

Therefore, the required probability is.

(ii) Let C be the event that at least one child is a girl.

The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).

An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

The given data can be tabulated as

True/False | Multiple choice | Total | |

Easy | 300 | 500 | 800 |

Difficult | 200 | 400 | 600 |

Total | 500 | 900 | 1400 |

Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions

Total number of questions = 1400

Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is

P (E ∩ M) =

Probability of selecting a multiple choice question, P (M), is

P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

∴

Therefore, the required probability is.

Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

When dice is thrown, number of observations in the sample space = 6 × 6 = 36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different. ∴ A = {(1, 3), (2, 2), (3, 1)}

B=1, 21, 31, 41, 51, 62, 12, 32, 42, 52,

63, 13, 23, 43, 53, 64, 14,

24, 34, 54, 65, 15, 25, 35, 45,66,

16, 26, 36, 46, 5 A∩B = 1, 3, 3, 1∴ P(B) =

3036=56and

PA∩B = 236=118Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

Therefore, the required probability is .

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event â€˜the coin shows a tailâ€, given that â€˜at least one die shows a 3â€.

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).

Therefore,

If

(A) 0 (B)

(C) not defined (D) 1

It is given that

Therefore, P (A|B) is not defined.

Thus, the correct answer is C.

If A and B are events such that P (A|B) = P(B|A), then

(A) A ⊂ B but A ≠ B (B) A = B

(C) A ∩ B = Φ (D) P(A) = P(B)

It is given that, P(A|B) = P(B|A)

⇒ P (A) = P (B)

Thus, the correct answer is D.

If, find P (A ∩ B) if A and B are independent events.

It is given that

A and B are independent events. Therefore,

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

There are 26 black cards in a deck of 52 cards.

Let P (A) be the probability of getting a black card in the first draw.

Let P (B) be the probability of getting a black card on the second draw.

Since the card is not replaced,

Thus, probability of getting both the cards black =

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.

Therefore, probability that first drawn orange is good, P (A)

The oranges are not replaced.

Therefore, probability of getting second orange good, P (B) =

Similarly, probability of getting third orange good, P(C)

The box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good

Therefore, the probability that the box is approved for sale is .

A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

Let A: Head appears on the coin

B: 3 on die

∴

Therefore, A and B are independent events.

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

When a die is thrown, the sample space (S) is

S = {1, 2, 3, 4, 5, 6}

Let A: the number is even = {2, 4, 6}

⇒

B: the number is red = {1, 2, 3}

⇒

PB =36=12∴ A ∩ B = {2}

Therefore, A and B are not independent.

Let E and F be events with. Are E and F independent?

It is given that, and

Therefore, E and F are not independent.

Given that the events A and B are such that P(A) =

12,

PA∪B=35and P (B) = *p*. Find *p* if they are (i) mutually exclusive (ii) independent.

It is given that

(i) When A and B are mutually exclusive, A ∩ B = *Φ*

∴ P (A ∩ B) = 0

It is known that,

(ii) When A and B are independent,

It is known that,

Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find

(i) P (A ∩ B) (ii) P (A ∪ B)

(iii) P (A|B) (iv) P (B|A)

It is given that P (A) = 0.3 and P (B) = 0.4

(i) If A and B are independent events, then

(ii) It is known that,

(iii) It is known that,

(iv) It is known that,

If A and B are two events such that, find P (not A and not B).

It is given that,

P(not on A and not on B) =

P(not on A and not on B) =

Events A and B are such that . State whether A and B are independent?

It is given that

Therefore, A and B are not independent events.

Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find

(i) P (A and B) (ii) P (A and not B)

(iii) P (A or B) (iv) P (neither A nor B)

It is given that P (A) = 0.3 and P (B) = 0.6

Also, A and B are independent events.

(i)

(ii) P (A and not B) =

(iii) P (A or B) =

(iv) P (neither A nor B) =

A die is tossed thrice. Find the probability of getting an odd number at least once.

Probability of getting an odd number in a single throw of a die =

Similarly, probability of getting an even number =

Probability of getting an even number three times =

Therefore, probability of getting an odd number at least once

= 1 − Probability of getting an odd number in none of the throws

= 1 − Probability of getting an even number thrice

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

(ii) first ball is black and second is red.

(iii) one of them is black and other is red.

Total number of balls = 18

Number of red balls = 8

Number of black balls = 10

(i) Probability of getting a red ball in the first draw =

The ball is replaced after the first draw.

∴ Probability of getting a red ball in the second draw =

Therefore, probability of getting both the balls red =

(ii) Probability of getting first ball black =

The ball is replaced after the first draw.

Probability of getting second ball as red =

Therefore, probability of getting first ball as black and second ball as red =

(iii) Probability of getting first ball as red =

The ball is replaced after the first draw.

Probability of getting second ball as black =

Therefore, probability of getting first ball as black and second ball as red =

Therefore, probability that one of them is black and other is red

= Probability of getting first ball black and second as red + Probability of getting first ball red and second ball black

Probability of solving specific problem independently by A and B arerespectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved (ii) exactly one of them solves the problem.

Probability of solving the problem by A, P (A) =

Probability of solving the problem by B, P (B) =

Since the problem is solved independently by A and B,

- Probability that the problem is solved = P (A ∪ B)

= P (A) + P (B) − P (AB)

(ii) Probability that exactly one of them solves the problem is given by,

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(i) E: ‘the card drawn is a spade’

F: ‘the card drawn is an ace’

(ii) E: ‘the card drawn is black’

F: ‘the card drawn is a king’

(iii) E: ‘the card drawn is a king or queen’

F: ‘the card drawn is a queen or jack’

(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces.

∴ P(E) = P(the card drawn is a spade) =

∴ P(F) = P(the card drawn is an ace) =

In the deck of cards, only 1 card is an ace of spades.

P(EF) = P(the card drawn is spade and an ace) =

P(E) × P(F) =

⇒ P(E) × P(F) = P(EF)

Therefore, the events E and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.

∴ P(E) = P(the card drawn is black) =

∴ P(F) = P(the card drawn is a king) =

In the pack of 52 cards, 2 cards are black as well as kings.

∴ P (EF) = P(the card drawn is a black king) =

P(E) × P(F) =

Therefore, the given events E and F are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.

∴ P(E) = P(the card drawn is a king or a queen) =

∴ P(F) = P(the card drawn is a queen or a jack) =

There are 4 cards which are king or queen and queen or jack.

∴ P(EF) = P(the card drawn is a king or a queen, or queen or a jack)

=

P(E) × P(F) =

Therefore, the given events E and F are not independent.

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English news papers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English news papers.

(b) If she reads Hindi news paper, find the probability that she reads English news paper.

(c) If she reads English news paper, find the probability that she reads Hindi news paper.

Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.

It is given that,

- Probability that a student reads Hindi or English newspaper is,

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P (E|H).

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (H|E).

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A) 0 (B) (C) (D)

When two dice are rolled, the number of outcomes is 36.

The only even prime number is 2.

Let E be the event of getting an even prime number on each die.

∴ E = {(2, 2)}

Therefore, the correct answer is D.

Two events A and B will be independent, if

(A) A and B are mutually exclusive

(B)

(C) P(A) = P(B)

(D) P(A) + P(B) = 1

Two events A and B are said to be independent, if P(AB) = P(A) × P(B)

Consider the result given in alternative **B**.

This implies that A and B are independent, if

**Distracter Rationale**

**A. **Let P (A) = *m*, P (B) = *n*, 0 < *m*, *n* < 1

A and B are mutually exclusive.

**C.** Let A: Event of getting an odd number on throw of a die = {1, 3, 5}

B: Event of getting an even number on throw of a die = {2, 4, 6}

Here,

**D.** From the above example, it can be seen that,

However, it cannot be inferred that A and B are independent.

Thus, the correct answer is B.

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

The urn contains 5 red and 5 black balls.

Let a red ball be drawn in the first attempt.

P (drawing a red ball)

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

P (drawing a red ball)

Let a black ball be drawn in the first attempt.

P (drawing a black ball in the first attempt)

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

P (drawing a red ball)

Therefore, probability of drawing second ball as red is

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Let E_{1} and E_{2} be the events of selecting first bag and second bag respectively.

Let A be the event of getting a red ball.

The probability of drawing a ball from the first bag, given that it is red, is given by P (E_{2}|A).

By using Bayes’ theorem, we obtain

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?

Let E_{1} and E_{2} be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.

The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by .

By using Bayes’ theorem, we obtain

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let be the probability that he knows the answer and be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability What is the probability that the student knows the answer given that he answered it correctly?

Let E_{1} and E_{2} be the respective events that the student knows the answer and he guesses the answer.

Let A be the event that the answer is correct.

The probability that the student answered correctly, given that he knows the answer, is 1.

∴ P (A|E_{1}) = 1

Probability that the student answered correctly, given that he guessed, is .

The probability that the student knows the answer, given that he answered it correctly, is given by .

By using Bayes’ theorem, we obtain

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Let E_{1} and E_{2} be the respective events that a person has a disease and a person has no disease.

Since E_{1} and E_{2} are events complimentary to each other,

∴ P (E_{1}) + P (E_{2}) = 1

⇒ P (E_{2}) = 1 − P (E_{1}) = 1 − 0.001 = 0.999

Let A be the event that the blood test result is positive.

Probability that a person has a disease, given that his test result is positive, is given by

P (E_{1}|A).

By using Bayes’ theorem, we obtain

There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Let E_{1}, E_{2}, and E_{3} be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.

Let A be the event that the coin shows heads.

A two-headed coin will always show heads.

Probability of heads coming up, given that it is a biased coin= 75%

Since the third coin is unbiased, the probability that it shows heads is always.

The probability that the coin is two-headed, given that it shows heads, is given by

P (E_{1}|A).

By using Bayes’ theorem, we obtain

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Let E_{1}, E_{2}, and E_{3} be the respective events that the driver is a scooter driver, a car driver, and a truck driver.

Let A be the event that the person meets with an accident.

There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.

Total number of drivers = 2000 + 4000 + 6000 = 12000

P (E_{1}) = P (driver is a scooter driver)

P (E_{2}) = P (driver is a car driver)

P (E_{3}) = P (driver is a truck driver)

The probability that the driver is a scooter driver, given that he met with an accident, is given by P (E_{1}|A).

By using Bayes’ theorem, we obtain

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that was produced by machine B?

Let E_{1} and E_{2} be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.

∴ Probability of items produced by machine A, P (E_{1})

Probability of items produced by machine B, P (E_{2})

Probability that machine A produced defective items, P (X|E_{1})

Probability that machine B produced defective items, P (X|E_{2})

The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E_{2}|X).

By using Bayes’ theorem, we obtain

Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Let E_{1} and E_{2} be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.

P (E_{1}) = Probability that the first group wins the competition = 0.6

P (E_{2}) = Probability that the second group wins the competition = 0.4

P (A|E_{1}) = Probability of introducing a new product if the first group wins = 0.7

P (A|E_{2}) = Probability of introducing a new product if the second group wins = 0.3

The probability that the new product is introduced by the second group is given by

P (E_{2}|A).

By using Bayes’ theorem, we obtain

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Let E_{1} be the event that the outcome on the die is 5 or 6 and E_{2} be the event that the outcome on the die is 1, 2, 3, or 4.

Let A be the event of getting exactly one head.

P (A|E_{1}) = Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6

P (A|E_{2}) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4

The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E_{2}|A).

By using Bayes’ theorem, we obtain

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?

Let E_{1}, E_{2}, and E_{3} be the respective events of the time consumed by machines A, B, and C for the job.

Let X be the event of producing defective items.

The probability that the defective item was produced by A is given by P (E_{1}|A).

By using Bayes’ theorem, we obtain

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Let E_{1} and E_{2 }be the respective events of choosing a diamond card and a card which is not diamond.

Let A denote the lost card.

Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

When one diamond card is lost, there are 12 diamond cards out of 51 cards.

Two cards can be drawn out of 12 diamond cards in ways.

Similarly, 2 diamond cards can be drawn out of 51 cards in ways. The probability of getting two cards, when one diamond card is lost, is given by P (A|E_{1}).

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.

Two cards can be drawn out of 13 diamond cards in ways whereas 2 cards can be drawn out of 51 cards in ways.

The probability of getting two cards, when one card is lost which is not diamond, is given by P (A|E_{2}).

The probability that the lost card is diamond is given by P (E_{1}|A).

By using Bayes’ theorem, we obtain

Probability that A speaks truth is. A coin is tossed. A reports that a head appears. The probability that actually there was head is

A.

B.

C.

D.

Let E_{1} and E_{2} be the events such that

E_{1}: A speaks truth

E_{2}: A speaks false

Let X be the event that a head appears.

P(E1)=45Therefore,

P(E2)=1-P(E1)=1-45=15If a coin is tossed, then it may result in either head (H) or tail (T).

The probability of getting a head is whether A speaks truth or not.

The probability that there is actually a head is given by P (E_{1}|X).

Therefore, the correct answer is A.

If A and B are two events such that A ⊂ B and P (B) ≠ 0, then which of the following is correct?

**A. **

**B. **

**C. **

**D. **None of these

If A ⊂ B, then A ∩ B = A

⇒ P (A ∩ B) = P (A)

Also, P (A) < P (B)

Consider … (1)

Consider … (2)

It is known that, P (B) ≤ 1

Thus, from (3), it can be concluded that the relation given in alternative C is correct.

State which of the following are **not **the probability distributions of a random variable. Give reasons for your answer.

**(i)**

X | 0 | 1 | 2 |

P (X) | 0.4 | 0.4 | 0.2 |

**(ii)**

X | 0 | 1 | 2 | 3 | 4 |

P (X) | 0.1 | 0.5 | 0.2 | − 0.1 | 0.3 |

**(iii)**

Y | −1 | 0 | 1 |

P (Y) | 0.6 | 0.1 | 0.2 |

**(iv)**

Z | 3 | 2 | 1 | 0 | −1 |

P (Z) | 0.3 | 0.2 | 0.4 | 0.1 | 0.05 |

It is known that the sum of all the probabilities in a probability distribution is one.

**(i)** Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1

Therefore, the given table is a probability distribution of random variables.

**(ii)** It can be seen that for X = 3, P (X) = −0.1

It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.

**(iii)** Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

**(iv)** Sum of the probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. What are the possible values of X? Is X a random variable?

The two balls selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball.

X represents the number of black balls.

∴X (BB) = 2

X (BR) = 1

X (RB) = 1

X (RR) = 0

Therefore, the possible values of X are 0, 1, and 2.

Yes, X is a random variable.

Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T)

X (5 H, 1 T)

X (4 H, 2 T)

X (3 H, 3 T)

X (2 H, 4 T)

X (1 H, 5 T)

X (0H, 6 T)

Thus, the possible values of X are 6, 4, 2, and 0.

Find the probability distribution of

**(i)** number of heads in two tosses of a coin

**(ii)** number of tails in the simultaneous tosses of three coins

**(iii)** number of heads in four tosses of a coin

**(i)** When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads.

∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

Therefore, X can take the value of 0, 1, or 2.

It is known that,

P (X = 0) = P (TT)

P (X = 1) = P (HT) + P (TH)

P (X = 2) = P (HH)

Thus, the required probability distribution is as follows.

**(ii)** When three coins are tossed simultaneously, the sample space is

Let X represent the number of tails.

It can be seen that X can take the value of 0, 1, 2, or 3.

P (X = 0) = P (HHH) =

P (X = 1) = P (HHT) + P (HTH) + P (THH) =

P (X = 2) = P (HTT) + P (THT) + P (TTH) =

P (X = 3) = P (TTT) =

Thus, the probability distribution is as follows.

**(iii)** When a coin is tossed four times, the sample space is

Let X be the random variable, which represents the number of heads.

It can be seen that X can take the value of 0, 1, 2, 3, or 4.

P (X = 0) = P (TTTT) =

P (X = 1) = P (TTTH) + P (TTHT) + P (THTT) + P (HTTT)

=

P (X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (HTTH) + P (HTHT)

+ P (THTH)

=

P (X = 3) = P (HHHT) + P (HHTH) + P (HTHH) P (THHH)

=

P (X = 4) = P (HHHH) =

Thus, the probability distribution is as follows.

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

**(i)** number greater than 4

**(ii)** six appears on at least one die

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

- Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) =

P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)

P (X = 2) = P (number greater than 4 on both the tosses)

Thus, the probability distribution is as follows.

**(ii)** Here, success means six appears on at least one die.

P (Y = 0 ) = P (six appears on none of the dice) =

56 × 56 = 2536P (Y = 1) = P (six appears on at least one of the dice) =

16 × 56 + 56 × 16 +16 × 16 =1136

Thus, the required probability distribution is as follows.

Y | 0 | 1 |

P (Y) | 2536 | 1136 |

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

P (X = 0) = P (4 non-defective and 0 defective)

P (X = 1) = P (3 non-defective and 1 defective)

P (X = 2) = P (2 non-defective and 2 defective)

P (X = 3) = P (1 non-defective and 3 defective)

P (X = 4) = P (0 non-defective and 4 defective)

Therefore, the required probability distribution is as follows.

A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Let the probability of getting a tail in the biased coin be *x*.

∴ P (T) =* x*

⇒ P (H) = 3*x*

For a biased coin, P (T) + P (H) = 1

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

Let X be the random variable representing the number of tails.

∴ P (X = 0) = P (no tail) = P (H) × P (H)

P (X = 1) = P (one tail) = P (HT) + P (TH)

P (X = 2) = P (two tails) = P (TT)

Therefore, the required probability distribution is as follows.

A random variable X has the following probability distribution.

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

P (X) | 0 | k | 2k | 2k | 3k | k^{2} | 2k^{2} | 7k^{2} + k |

Determine

**(i)** *k*

**(ii)** P (X < 3)

**(iii)** P (X > 6)

**(iv)** P (0 < X < 3)

**(i) **It is known that the sum of probabilities of a probability distribution of random variables is one.

*k* = − 1 is not possible as the probability of an event is never negative.

∴

**(ii)** P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

**(iii)** P (X > 6) = P (X = 7)

**(iv)** P (0 < X < 3) = P (X = 1) + P (X = 2)

The random variable X has probability distribution P(X) of the following form, where *k* is some number:

(a) Determine the value of *k*.

(b) Find P(X < 2), P(X ≥ 2), P(X ≥ 2).

(a) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴ *k* + 2*k* + 3*k* + 0 = 1

⇒ 6*k* = 1

⇒ *k* =

(b) P(X < 2) = P(X = 0) + P(X = 1)

Find the mean number of heads in three tosses of a fair coin.

Let X denote the success of getting heads.

Therefore, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0, 1, 2, or 3.

∴ P (X = 1) = P (HHT) + P (HTH) + P (THH)

∴P(X = 2) = P (HHT) + P (HTH) + P (THH)

Therefore, the required probability distribution is as follows.

Mean of X E(X), µ =

Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) =

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

P (X = 2) = P (six on both the dice) =

Therefore, the required probability distribution is as follows.

Then, expectation of X = E(X) =

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denotes the larger of the two numbers obtained. Find E(X).

The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6,5) , (6, 4), (6, 3), (6, 2), and (6, 1).

Therefore, the required probability distribution is as follows.

Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) =

P(X = 3) = P (1, 2) + P(2, 1) =

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) =

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) =

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) =

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1)

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) =

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) =

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) =

P(X = 11) = P(5, 6) + P(6, 5) =

P(X = 12) = P(6, 6) =

Therefore, the required probability distribution is as follows.

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is.

The given information can be compiled in the frequency table as follows.

X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |

f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |

P(X = 14) =, P(X = 15) =, P(X = 16) =, P(X = 16) =,

P(X = 18) =, P(X = 19) =, P(X = 20) =, P(X = 21) =

Therefore, the probability distribution of random variable X is as follows.

Then, mean of X = E(X)

E(X^{2}) =

In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).

It is given that P(X = 0) = 30% =

Therefore, the probability distribution is as follows.

X | 0 | 1 |

P(X) | 0.3 | 0.7 |

It is known that, Var (X) =

= 0.7 − (0.7)^{2}

= 0.7 − 0.49

= 0.21

The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is

(A) 1 (B) 2 (C) 5 (D)

Let X be the random variable representing a number on the die.

The total number of observations is six.

Therefore, the probability distribution is as follows.

Mean = E(X) =

The correct answer is B.

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is

(A) (B) (C) (D)

Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

∴ P (X = 0) = P (0 ace and 2 non-ace cards) =

P (X = 1) = P (1 ace and 1 non-ace cards) =

P (X = 2) = P (2 ace and 0 non- ace cards) =

Thus, the probability distribution is as follows.

Then, E(X) =

Therefore, the correct answer is D.

A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of

(i) 5 successes? (ii) at least 5 successes?

(iii) at most 5 successes?

The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is,

X has a binomial distribution.

Therefore, P (X = *x*) =

(i) P (5 successes) = P (X = 5)

(ii) P(at least 5 successes) = P(X ≥ 5)

(iii) P (at most 5 successes) = P(X ≤ 5)

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.

The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

Clearly, X has the binomial distribution with *n* = 4,

∴ P (2 successes) = P (X = 2)

There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Let X denote the number of defective items in a sample of 10 items drawn successively. Since the drawing is done with replacement, the trials are Bernoulli trials.

X has a binomial distribution with *n* = 10 and

P(X = *x*) =

P (not more than 1 defective item) = P (X ≤ 1)

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

(i) all the five cards are spades?

(ii) only 3 cards are spades?

(iii) none is a spade?

Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.

In a well shuffled deck of 52 cards, there are 13 spade cards.

X has a binomial distribution with *n* = 5 and

(i) P (all five cards are spades) = P(X = 5)

(ii) P (only 3 cards are spades) = P(X = 3)

(iii) P (none is a spade) = P(X = 0)

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is the probability that out of 5 such bulbs

(i) none

(ii) not more than one

(iii) more than one

(iv) at least one

will fuse after 150 days of use.

Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, *p* = 0.05

X has a binomial distribution with *n* = 5 and *p* = 0.05

(i) P (none) = P(X = 0)

(ii) P (not more than one) = P(X ≤ 1)

(iii) P (more than 1) = P(X > 1)

(iv) P (at least one) = P(X ≥ 1)

A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

Since the balls are drawn with replacement, the trials are Bernoulli trials.

X has a binomial distribution with *n* = 4 and

P (none marked with 0) = P (X = 0)

In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Let X represent the number of correctly answered questions out of 20 questions.

The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.

∴ *p* =

X has a binomial distribution with *n *= 20 and *p* =

P (at least 12 questions answered correctly) = P(X ≥ 12)

Suppose X has a binomial distribution. Show that X = 3 is the most likely outcome.

(Hint: P(X = 3) is the maximum among all P (*x*_{i}), *x*_{i} = 0, 1, 2, 3, 4, 5, 6)

X is the random variable whose binomial distribution is.

Therefore, *n* = 6 and

It can be seen that P(X = *x*) will be maximum, if will be maximum.

The value ofis maximum. Therefore, for *x* = 3, P(X = x) is maximum.

Thus, X = 3 is the most likely outcome.

On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.

Probability of getting a correct answer is, *p*

Clearly, X has a binomial distribution with *n* = 5 and *p*

P (guessing more than 4 correct answers) = P(X ≥ 4)

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is. What is the probability that he will in a prize (a) at least once (b) exactly once (c) at least twice?

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with *n* = 50 and

(a) P (winning at least once) = P (X ≥ 1)

(b) P (winning exactly once) = P(X = 1)

(c) P (at least twice) = P(X ≥ 2)

Find the probability of getting 5 exactly twice in 7 throws of a die.

The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.

Probability of getting 5 in a single throw of the die, *p *

Clearly, X has the probability distribution with *n* = 7 and *p *

P (getting 5 exactly twice) = P(X = 2)

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die,* p*

Clearly, X has a binomial distribution with *n* = 6

P (at most 2 sixes) = P(X ≤ 2)

It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.

Clearly, X has a binomial distribution with *n* = 12 and *p *= 10% =

P (selecting 9 defective articles) =

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

**(A)** 10^{−1}

**(B)**

**(C)**

**(D)**

The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.

Probability of getting a defective bulb, *p*

Clearly, X has a binomial distribution with *n* = 5 and

P (none of the bulbs is defective) = P(X = 0)

The correct answer is C.

The probability that a student is not a swimmer is. Then the probability that out of five students, four are swimmers is

**(A)** **(B)**

**(C)** **(D)** None of these

The repeated selection of students who are swimmers are Bernoulli trials. Let X denote the number of students, out of 5 students, who are swimmers.

Probability of students who are not swimmers, *q*

Clearly, X has a binomial distribution with *n* = 5 and

P (four students are swimmers) = P(X = 4)

Therefore, the correct answer is A.

A and B are two events such that P (A) ≠ 0. Find P (B|A), if

(i) A is a subset of B (ii) A ∩ B = Φ

It is given that, P (A) ≠ 0

(i) A is a subset of B.

(ii)

A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

(ii) Find the probability that both children are females, if it is known that the elder child is a female.

If a couple has two children, then the sample space is

S = {(*b*, *b*), (*b*, *g*), (*g*, *b*), (*g*, *g*)}

(i) Let E and F respectively denote the events that both children are males and at least one of the children is a male.

(ii) Let A and B respectively denote the events that both children are females and the elder child is a female.

Suppose that 5% of men and 0.25% of women have grey hair. A haired person is selected at random. What is the probability of this person being male?

Assume that there are equal number of males and females.

It is given that 5% of men and 0.25% of women have grey hair.

Therefore, percentage of people with grey hair = (5 + 0.25) % = 5.25%

∴ Probability that the selected haired person is a male

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

A person can be either right-handed or left-handed.

It is given that 90% of the people are right-handed.

Using binomial distribution, the probability that more than 6 people are right-handed is given by,

Therefore, the probability that at most 6 people are right-handed

= 1 − P (more than 6 are right-handed)

An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

(i) all will bear ‘X’ mark.

(ii) not more than 2 will bear ‘Y’ mark.

(iii) at least one ball will bear ‘Y’ mark

(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

*p* = P (ball bearing mark ‘X’) =

*q* = P (ball bearing mark ‘Y’) =

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with *n* = 6 and *p* =.

∴ P (Z = *z*) =

(i) P (all will bear ‘X’ mark) = P (Z = 0) =

(ii) P (not more than 2 bear ‘Y’ mark) = P (Z ≤ 2)

= P (Z = 0) + P (Z = 1) + P (Z = 2)

(iii) P (at least one ball bears ‘Y’ mark) = P (Z ≥ 1) = 1 − P (Z = 0)

(iv) P (equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z = 3)

=

C36 253 353

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is. What is the probability that he will knock down fewer than 2 hurdles?

Let *p* and *q* respectively be the probabilities that the player will clear and knock down the hurdle.

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Therefore, by binomial distribution, we obtain

P (X = *x*) =

P (player knocking down less than 2 hurdles) = P (X < 2)

= P (X = 0) + P (X = 1)

=

A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

The probability of getting a six in a throw of die isand not getting a six is.

Let

The probability that the 2 sixes come in the first five throws of the die is

∴ Probability that third six comes in the sixth throw =

If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

In a leap year, there are 366 days i.e., 52 weeks and 2 days.

In 52 weeks, there are 52 Tuesdays.

Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining 2 days will be Tuesdays.

The remaining 2 days can be

Monday and Tuesday

Tuesday and Wednesday

Wednesday and Thursday

Thursday and Friday

Friday and Saturday

Saturday and Sunday

Sunday and Monday

Total number of cases = 7

Favourable cases = 2

∴Probability that a leap year will have 53 Tuesdays =

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

The probability of success is twice the probability of failure.

Let the probability of failure be *x*.

∴ Probability of success = 2*x*

Let *p* = and *q* =

Let X be the random variable that represents the number of successes in six trials.

By binomial distribution, we obtain

P (X = *x*) =

Probability of at least 4 successes = P (X ≥ 4)

= P (X = 4) + P (X = 5) + P (X = 6)

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Let the man toss the coin *n* times. The *n* tosses are *n* Bernoulli trials.

Probability (*p*) of getting a head at the toss of a coin is.

∴ *p* = ⇒ *q* =

It is given that,

P (getting at least one head) >

P (*x* ≥ 1) > 0.9

⇒ 1 − P (*x* = 0) > 0.9

The minimum value of *n* that satisfies the given inequality is 4.

Thus, the man should toss the coin 4 or more than 4 times.

In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

When a die is thrown, then probability of getting a six =

16then, probability of not getting a six =

1 – 16 = 56If the man gets a six in the first throw, then

probability of getting a six =

16If he does not get a six in first throw, but gets a six in second throw, then

probability of getting a six in the second throw =

56×16 = 536If he does not get a six in the first two throws, but gets in the third throw, then

probability of getting a six in the third throw =

56×56×16 = 25216probability that he does not get a six in any of the three throws =

56×56×56 = 125216In the first throw he gets a six, then he will receive Re 1.

If he gets a six in the second throw, then he will receive Re (1 – 1) = 0

If he gets a six in the third throw, then he will receive Rs(-1 – 1 + 1) = Rs (-1), that means he will lose Re 1 in this case.

Expected value =

16×1 + 56×16 × 0 + 56×56×16×-1 = 11216So, he will loose Rs

11216.

Suppose we have four boxes. A, B, C and D containing coloured marbles as given below:

Box | Marble colour | ||

Red | White | Black | |

A | 1 | 6 | 3 |

B | 6 | 2 | 2 |

C | 8 | 1 | 1 |

D | 0 | 6 | 4 |

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Let R be the event of drawing the red marble.

Let E_{A}, E_{B}, and E_{C} respectively denote the events of selecting the box A, B, and C.

Total number of marbles = 40

Number of red marbles = 15

Probability of drawing the red marble from box A is given by P (E_{A}|R).

Probability that the red marble is from box B is P (E_{B}|R).

Probability that the red marble is from box C is P (E_{C}|R).

Assume that the chances of the patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Let A, E_{1}, and E_{2} respectively denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.

Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by P (E_{1}|A).

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability).

The total number of determinants of second order with each element being 0 or 1 is (2)^{4} = 16

The value of determinant is positive in the following cases.

∴ Required probability =

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = 0.2

P(B fails alone) = 0.15

P(A and B fail) = 0.15

Evaluate the following probabilities

(i) P(A fails| B has failed) (ii) P(A fails alone)

Let the event in which A fails and B fails be denoted by E_{A} and E_{B}.

P (E_{A}) = 0.2

P (E_{A }∩ E_{B}) = 0.15

P (B fails alone) = P (E_{B}) − P (E_{A} ∩ E_{B})

⇒ 0.15 = P (E_{B}) − 0.15

⇒ P (E_{B}) = 0.3

(i)

(ii) P (A fails alone) = P (E_{A}) − P (E_{A }∩ E_{B})

= 0.2 − 0.15

= 0.05

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Let E_{1} and E_{2} respectively denote the events that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

When a black ball is transferred from bag I to II,

If A and B are two events such that P (A) ≠ 0 and P(B|A) = 1, then.

**(A)** A ⊂ B

**(B)** B ⊂ A

**(C)** B = Φ

**(D)** A = Φ

P (A) ≠ 0 and

Thus, the correct answer is A.

If P (A|B) > P (A), then which of the following is correct:

**(A)** P (B|A) < P (B) **(B)** P (A ∩ B) < P (A).P (B)

**(C)** P (B|A) > P (B) **(D)** P (B|A) = P (B)

Thus, the correct answer is C.

If A and B are any two events such that P (A) + P (B) − P (A and B) = P (A), then

**(A)** P (B|A) = 1 **(B)** P (A|B) = 1

**(C)** P (B|A) = 0 **(D)** P (A|B) = 0

Thus, the correct answer is B.

**Conclusion**: If you liked NCERT Class 12 Maths Solutions Chapter 13 – Probability, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Maths Solutions Chapter 13 – Probability appeared first on Babaji Academy.

]]>The post NCERT Class 12 Maths Solutions Chapter 12 – Linear Programming appeared first on Babaji Academy.

]]>In addition to this, you can also check our previous post, we have updated solution of Chapter 11 i.e Three Dimensional Geometry. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 1

Solution 2

Let the farmer mix x bags of brand P and y bags of brand Q.

The given information can be compiled in a table as follows.

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

**Conclusion**: If you liked NCERT Class 12 Maths Solutions Chapter 12 – Linear Programming, then share it with others. You can also share your feedback on this post in the comment box.

The post NCERT Class 12 Maths Solutions Chapter 12 – Linear Programming appeared first on Babaji Academy.

]]>