Monthly Archives: January 2022

In This Post we are  providing Chapter- 13 NUCLEI NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON NUCLEI

Question 1.
Calculate the binding energy per nucleon of Fe⁵⁶₂₆ Given m_Fe = 55.934939 u, m_n = 1.008665 u and m_p = 1.007825 u
Answer:
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z m_p + (A – Z)m_n – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.
Answer:

Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Answer:
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 10²³ nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10^-3 g are

Question 4.
The half-life of 238 92U is 4.5 × 10⁹ years. Calculate the activity of 1 g sample of ₉₂²³⁸U.
Answer:
Given T = 4.5 × 10⁹ years.
Number of nuclei of U in 1 g
= N = 6.023×1023238 = 2.5 × 10²¹

Therefore activity

Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Answer:
Given λ = 0.3456 day^-1
or
T_1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = N_oe^-λt
or
N = N_o e^{-0 3456 × 4}
or
N = N₀ e^{-1 3824} = N_o × 0.25

Therefore the percentage of undecayed is

Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Answer:
Given N/N_o = 6.25 %, t = 16 days, λ = ?

Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day^-1

Question 7.
A radioactive substance decays to 1/32^th of its initial value in 25 days. Calculate its half-life.
Answer:
Given t = 25 days, N = N_o / 32, using

Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Answer:
Given T₁/2 = 30 s, N = 3N_o / 4, λ = ?, t = ?
(i) Decay constant
λ = 0.693T1/2=0.69330 = 0.0231 s^-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Answer:
Using N = N_oe^-λt we have

Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
Answer:
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given N_ox = N_oY, T_X = 1 h, T_Y = 2 h, therefore
λXλY=21 = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are R_x and R_y respectively, then

Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.

The mass of the star is 5 × 10³² kg and it generates energy at the rate of 5 × 10³⁰ watt. How long will it take to convert all the helium to carbon at this rate?

Answer:

As 4 × 10^-3 kg of He consists of 6.023 × 10²³ He nuclei so 5 × 10³² kg He will contain
6.023×1023×5×10324×10−3 = 7.5 × 10⁵⁸ nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10^-13 J of energy
So all nuclei in the star will produce
E = 7.27×1.6×10−133 × 7.5 × 10⁵⁸
= 2.9 × 10⁴⁶ J

As power generated is P = 5 × 10³⁰ W, therefore time taken to convert all He nuclei into carbon is
t = EP=2.9×10465×1030 = 5.84 × 10¹⁵ s
or
1.85 × 10⁸ years

In This Post we are  providing Chapter-4 MOVING CHARGES AND MAGNETISM NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON MOVING CHARGES AND MAGNETISM

Question 1.
Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, stating clearly the function of the electric and magnetic field applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle.
Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

Question 2.
Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.
(i) Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path.
(ii) What is resonance condition? How is it used to accelerate the charged particles? (All India 2017)
Answer:
(i) Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

(ii) The frequency v_a of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one-half of the revolution. The requirement v_a = v_c is called the resonance condition.
The phase of the supply is adjusted so that when the positive ions arrive at the edge of D₁, D₂ is at a lower potential and the ions are accelerated across the gap.

Question 3.
(a) Two straight long parallel conductors carry currents I₁ and I₂ in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them.
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.
(i) What is the direction of the magnetic moment of the current loop?
(ii) When is the torque acting on the loop
(A) maximum,
(B) zero?

Answer:
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A


(i) Magnetic moment will be out of the plane from the surface HEFG.
(ii) Torque
(A) Torque is maximum when MII B i.e., when it gets rotated by 90°.
(B) Torque is minimum when M and B are at 270° to each other.

Question 4.
(a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.
(b) What is the importance of a radial magnetic field and how is it produced?
(c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used? (All India)
Answer:
(a) Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB

The spring S_p attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring

In equilibrium state,

Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.

(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.

(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.

…where [R = Resistance of the coil

does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) For radial magnetic field, sin θ = 1,
so torque τ = NIAB.
Thus when radial magnetic field is used, the deflection of the coil is proportional to the current flowing through it. Hence a linear scale can be used to determine the deflection of the coil.

(c) A high resistance is joined in series with a galvanometer so that when the arrangement (voltmeter) is used in parallel with the selected section of the circuit, it should draw least amount of current. In case voltmeter draws appreciable amount of current, it will disturb the original value of potential difference by a good amount.

To convert a galvanometer into ammeter, a shunt is used in parallel with it so that when the arrangement is joined in series, the maximum current flows through the shunt, and thus the galvanometer is saved from its damage, when the current is passed through ammeter.

Question 5.
(a) Derive an expression for the force between two long parallel current carrying conductors.

(b) Use this expression to define S.I. unit of current.
(c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?
Answer:
(a) For (a) and (b) :
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A


(c) Force experienced by the proton,

As magnetic field due to the current carrying wire is directed into the plane of the paper (θ = 90°)

Force is directed away from the current carrying wire or in the right direction of observer.

Question 6.
State Biot-Savart law, giving the mathematical expression for it.
Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.
How does a circular loop carrying current behave as a magnet? (Delhi 2011)
Answer:
According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.


Magnetic field on the axis of circular current loop :

As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain

In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

Question 7.
With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles. Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.

Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

Yes, there is an upper limit. The increase in the kinetic energy of particles is qv. Therefore, the radius of their path goes on increasing each time, their kinetic energy increases. The lines are repeatedly accelerated across the dees, untill they have the required energy to have a radius approximately that of the dees. Hence, this is the upper limit on the energy required by the particles due to definite size of dees.

Question 8.
(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
(b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Justify this statement
(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range. (All India 2011)
Answer:
(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB

The spring S_p attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring

In equilibrium state,

Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.

(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.

…where [R = Resistance of the coil

does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Since vs=IsR increase in current sensitivity may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(c) Conversion of galvanometer into ammeter: By just connecting a low resistance known as shunt in parallel to the galvanometer, it can be converted into an ammeter.

Let G = resistance of the galvanometer.
I_g = the current with which galvanometer gives full scale deflection.
S = shunt resistance
I – I_g = current through the shunt.
As the galvanometer and shunt are connected in parallel,
Potential difference across the galvanometer = Potential difference across the shunt

Question 9.
(a) Write the expression for the force, F→, acting on a charged particle of charge ‘q’, moving with a velocity latex]\overrightarrow{\mathbf{v}}[/latex] in the presence of both electric field E→ and magnetic field B→. Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field B→. Prove that the torque τ⃗  acting on the loop is given by τ⃗ =m⃗ ×B→, where m→ is the magnetic moment of the loop. (All India 2011)
Answer:
(a) A charge q in an electric field E→ experiences the electric force, F→e=qE→

This force acts in the direction of field E→ and is independent of the velocity of the charge.

The magnetic force experienced by the charge q moving with velocity v→ in the magnetic field B is given by

This force acts perpendicular to the plane of V→ and B→ and depends on the velocity v→ of the charge.

The total force, or the Lorentz force, experienced by the charge q due to both electric and magnetic field is given by

Hence, A stationary charged particle does not experience any force in a magnetic field. (b) Torque on a current loop in a uniform magnetic field.
Let I = Current flowing through the coil PQRS

Its magnitude is, F3 = IaB sin(90° + 0)

Question 10.
(a) Explain, giving reasons, the basic difference in converting a galvanometer into
(i) a voltmeter and
(ii) an ammeter.
(b) Two long straight parallel conductors carrying steady currents I₁ and I₂ are separated by a distance’d’ Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
Answer:
(a) (i) Voltmeter is connected in parallel with the circuit element across which the potential difference is intended to be measured.

A galvanometer can be converted into a voltmeter by connecting a higher resistance in series with it. The value of this resistance is so adjusted that only current I which produces full scale deflection in the galvanometer, passes through the galvanometer.

(ii) A galvanometer can be converted into an ammeter by connecting a low value
resistance in parallel with it.


(b)
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A

In This Post we are  providing Chapter- 7 ALTERNATING CURRENT NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ALTERNATING CURRENT

Question 1.
The figure shows how the reactance of an Inductor varies with frequency. (a) Calculate the value of the Inductance of the Inductor using Information given In the graph. (b) If this Inductor is connected In senes to a resistor of 8 ohms, find what would be the impedance at 300 Hz.

Answer:
(a) We know that X_L = 2πfL or L = XL2πf.
Now slope of the graph is
XLf=8−6400−300=2100 = 0.02

Therefore L is L = XL2πf=0.022×3.14 = 0.0032 H

(b) NowR = 8 ohm, f = 300 Hz, Z = ?
Now Z = R2+X2L−−−−−−−√ . Therefore we have
Z = R2+X2L−−−−−−−√ = (8)2+(6)2−−−−−−−−−√ = 10 Ω

Question 2.
A 25.0 μF capacitor, a 0.10-henry Inductor, and a 25.0-ohm resistor are connected in series with an ac source whose emf is E= 310 s. In 314 t (i) what is the frequency of the .mf? (ii) Calculate (a) the reactance of the circuit (b) the Impedance of the circuit and (C) the current in the circuit.
Answer:
Given C = 25.0 μF, L = 0.10 henry, R = 25.0 ohm, E_o = 310V,
Comparing with the equation E = E_o sin ωt,
we have
(i) ω = 314 or f = 50 Hz

Question 3.
A sinusoidal voltage V = 200 sin 314 t Is applied to a resistor of 10 ohms. Calculate (i) rms value of current (ii) rms value of voltage and (iii) power dissipated as heat in watt.
Answer:
V_o = 200 V, ω = 314 rads^-1 , V_rms = ?, l_rms = ?, P = ?

Question 4.
Find the inductance of the inductor used in series with a bulb of resistance 10 ohms connected to an ac source of 80 V, 50 Hz. The power factor of the circuit is 0.5. Also, calculate the power dissipation in the circuit.
Answer:
Given R = 10 ohm, V = 80 Hz, f = 50 Hz, cos Φ = 0.5 , L = ?, P = ?
Using the formula
cos Φ = RR2+X2L√
or
0.25(R² + X²) = R²
Or
O.25X_L² = O.75R²
or
X_L = 17.32 ohm

Now using X_L = 2πfL we have
L = XL2πf=17.322×3.14×50 = 0.055 H

Now Z = R2+X2L−−−−−−−√
= (10)2+(17.32)2−−−−−−−−−−−−−√ = 20 Ω

Now P_av = l_rms V_rms cos Φ = V2mZ × cos Φ
or
Pav = (80)22×20 × 0.5 = 160 W

Question 5.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and Is in phase with the applied voltage. When the same voltage is applied across a device Y, the same current again flows through it, but it leads the voltage by π/2. If element ‘X’ is a pure resistor of 100 ohms,
(a) name the circuit element ‘Y’ and
Answer:
The element Y is a capacitor.

(b) calculate the rms value of current, if rms value of voltage is 141 V.
Answer:
The value of Xc is obtained as below

X_C = VI=2200.5 = 440 ohm

Therefore impedance of the circuit
Z = R2+X2C−−−−−−−√=(100)2+(440)2−−−−−−−−−−−−√ = 451.2 ohm

Therefore rms value of current V 141
l = VrmsZ=141451.2 = 0.3125 A

Question 6.
When an alternating voltage of 220 V is applied across a device X, a current of 0. 5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across a device V, the same current again flows through it, but it lags the voltage by π/2.
(a) Name the devices X and Y.
Answer:
The element X is a resistor and Y is an inductor.

(b) Calculate the current flowing through the circuit when the same voltage is applied across the series combination of the two devices X and Y.
Answer:
Now both R and XL are the same and are given by
R = X_L = 2200.5 = 440 ohm

Hence impedance of the circuit
Z = R2+X2L−−−−−−−√
= (440)2+(440)2−−−−−−−−−−−−√
= 622.2 ohm

Therefore current flowing through the circuit is
l = VZ = 220622.2 = 0.353 A

Question 7.
A 15.0 μF capacitor Is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) In the circuit. If the frequency Is doubled, what happens to the capacitive reactance and the current?
Answer:
Given C= 15.0 μF= 15 × 10^-6 F, V= 220 V,
f = 50 HZ, X_C = ? l_m = ?

The capacitive reactance is

Now l_rms = VrmsXc=220212 = 1.04 A

Peak value of current
l_m = 2–√ × lrms = 1.4.1 × 1.04 = 1.47 A

This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 8.
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Answer:
Given V_m = 283 V, f = 50 Hz, R = 3 Ω
L = 25.48 mH, and C = 796 μF.
(a) To find the impedance of the circuit, we first calculate XL and Xc
X_L = 2πfL = 2 × 3.14 × 50 × 25.48 × 10^-3 = 8 Ω
X_C = 12πfC
= 12×3.14×50×796×10−6 = 4 Ω

Therefore impedance of the circuit is
Z = R2+(XL−XC)2−−−−−−−−−−−−−−√
= 32+(8−4)2−−−−−−−−−−√
= 5 Ω

(b) Phase difference
Φ = tan^-1XL−XcR = tan-1 8−43 = 53.1°
Since Φ is positive therefore voltage leads current by the above phase.

(c) The power dissipated in the circuit is
P = V2rmsZ=V2m2Z=(283)22×5 = 8008.9 W
(d) power factor = cos Φ = cos 53.1° = 0.6

Question 9.
A capacitor and a resistor are connected in series with an ac source. If the potential difference across the C, R is 120 V and 90 V respectively, and if rms current of the circuit is 3 A, calculate the (i) impedance and (ii) power factor of the circuit.
Answer:
Given V_C = 120 V, V_R = 90 V, f = 3 A. and R = 90/3 = 30 ohm ,

Effective voltage in the circuit
V = V2C+V2R−−−−−−−√=(120)2+(90)2−−−−−−−−−−−√= 150 V .
(i) Therefore impedance of the circuit
Z = Vl=1503 = 50 Ω.

(ii) Now power factor of the circuit is
cos Φ = RZ=3050 = 0.6

Question 10.
An inductor 200 mH, a capacitor C, and a resistor 10 ohm are connected in series with 100 V, 50 Hz ac source. If the current and the voltage are in phase with each other, calculate the capacitance of the capacitor.
Answer:
When current and voltage are in phase then X_L = X_C

In This Post we are  providing Chapter-9 RAY OPTICS AND OPTICAL INSTRUMENTS NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON RAY OPTICS AND OPTICAL INSTRUMENT

Question 1.
A convex lens made up of a glass of refractive index 1.5 is dipped, In turn, In
(a) a medium of refractive index 1.65,
Answer:
When dipped in the medium of refractive index 1.65, it will behave as a concave lens and when dipped in the medium of refractive index 1.33, it will behave as a convex lens.

(b) a medium of refractive index 1.33.
(i) Will it behave as a converging or a diverging lens in the two cases?
Answer:
Its focal length in another medium is given by

Thus f_m = -5.5 f_a, i.e. focal length increases and becomes negative.

(ii) How will Its focal length change In the two media? (CBSE AI 2011)
Answer:
Similarly

Thus f_m = 3.3 f_a, i.e. focal length increases.

Question 2.
A compound microscope uses an objective lens of focal length 4 cm and an eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also, calculate the length of the microscope. (CBSE Al 2011)
Answer:
f_o = 4 cm, f_e = 10 cm, u_o = – 6 cm, M = ?, L = ?
Using

Hence angular magnification


Question 3.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 10⁶ m and the radius of the lunar orbit is 3.8 × 10⁸ m. (CBSE AI 2011, Delhi 2015)
Answer:
Given f_o = 15 m, f_e = 1.0 cm = 0.01 m, M = ?
Dm = 3.48 × 10⁶ m, r = 3.8 × 10⁸ m,

Using M = f0fe=150.01 = 1500

The angle subtended by the moon at the objective of the telescope

Question 4.
A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. Draw a ray diagram and find the location of the point at which the beam would now converge. (CBSE Delhi 2011C)
Answer:
The ray diagram is shown in the figure. In the absence of the concave lens the beam converges at point P. When the concave lens is introduced, the incident beam of light is diverged and now comes to focus at point Q. Thus for the concave lens P serves as a virtual object giving rise to a real Image at Q.

Here u = + 12 cm, f = – 16 cm, v = ?, Now for a lens

Hence v = 48 cm
i. e. the point at which the beam is focused is 48 cm from the lens.

Question 5.
Two convex lenses of focal length 20 cm and 1 cm constitute a telescope. The telescope is focused on a point that is 1 m away from the objective. Calculate the magnification produced and the length of the tube, if the final image Is formed at a distance of 25 cm from the eyepiece.
Answer:
Given f_o = 2o cm, f_e = 1 cm, u = – 100 cm, M =?, y =?

Fora lens 1v−1u=1f
or
1v−1−100=120
or
v = 25cm

Since the eye lens forms the image of the virtual object at the distance of distinct vision for the eye lens
v = – 25 cm, f_e = 1 cm,

Now 1v−1u=1f
or
1−25−1u = 1
or
u = – 2526cm

Now magnification produced by the object lens
m_o = vu=−25100=−14

Magnification produced by the eye Lens
m_e = vu=−25−25 × 26 = 26

Hence total magnification
M = m_o × m_e = -1 /4 × 26 = – 6.5

Question 6.
(a) Under what conditions are the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.
(b) Three lenses of focal lengths +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

Answer:
(a) (i) Light travels from a denser medium to a rarer medium.
(ii) Angle of Incidence in the denser medium is more than the critical angle for a given pair of media.
For the grazing incidence n sin i_C = l sin 90°
n = 1sinic

(b) For convex lens f = + 10 cm

Object distance for concave lens u₂ = 15 – 5 = 10 cm

For third lens
1f3=1v3−1∞ ⇒ v₃ = 30 cm

Question 7.
A ray of light incident on an equilateral glass prism (μ_g = 3–√) moves parallel to the baseline of the prism inside it. Find the angle of Incidence for this ray.
Answer:
Given A = 60°, μ_g = 3–√, i = ?

Using the expression μ = sinisinA/2
or

Question 8.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.

Answer:
The critical angle for the two rays is

This shows that the angle of Incidence for ray ‘2’ Is greater than the critical angle. Hence it suffers total internal reflection, white ray ‘1’ does not. Hence the path of rays is as shown.

Question 9.
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed. (CBSE AI 2014)
Answer:
The ray diagram is as shown.

For the convex lens, we have
u₁ = – 60 cm, f = + 20 cm, v = ?

Using lens formula we have

Had there been only the Lens, the image would have been formed at Q₁, which acts as a virtual object for the convex mirror.
Therefore u₂ = OQ₁ – OO’ = 30 – 15 = 15 cm

Using mirror formuLa we have
1v2+1u2=2R
or
1v2+1u15=220

Solving for v₂ we have
v₂ = 30cm

Hence the final image is formed at (Point Q) a distance of 30 cm behind the mirror.

Question 10.
A ray PQ is an incident normally on the face AB of a triangular prism refracting angle of 60°, made of a transparent material of refractive index 2 / 3–√, as shown in the figure. Trace the path of the ray as it passes through the prism. Also, calculate the angle of emergence and angle deviation.

Answer:
Critical angle for glass
µ = 1sinic
or
sin i_c = 1μ=3√2= 0.866
or
i_c = 60°

Now the ray is incident at an angle of 60° which is equal to the critical angle, therefore the ray graces the other edge of the prism

Therefore the angle of emergence is = 90°
Hence δ = 30°

In This Post we are  providing Chapter- 8 ELECTROMAGNETIC WAVESNCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROMAGNETIC WAVES

Question 1.
Answer the following:
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
Answer:
Gamma rays.
Frequency range > 3 × 10²⁰ Hz

(b) Thin ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race.

(c) An em wave exerts pressure on the surface on which it is incident. Justify. (CBSE Delhi 2014)
Answer:
An em wave carries a linear momentum with it. The linear momentum carried by a portion of a wave having energy U is given by p = U/c.

Thus, if the wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface. If the wave is totally reflected, the momentum delivered is p = 2U/c because the momentum of the wave changes from p to – p. Therefore, it follows that an em wave incident on a surface exerts a force and hence a pressure on the surface.

Question 2.
Answer the following questions:
(a) Why is the thin ozone layer at the top of the stratosphere crucial for human survival? Identify to which part of the electromagnetic spectrum does this radiation belongs and write one important application of the radiation.
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race. This radiation is UV radiation. It is used in sterilization.

(b) Why are infrared waves referred to as heat rays? How are they produced? What role do they play in maintaining the earth’s warmth through the greenhouse effect?
Answer:
Infrared radiations heat up the material on which they fall, hence they are also called heat rays. They are produced by the vibration of atoms and molecules. After falling on the earth, they are reflected back into the earth’s atmosphere. The earth’s atmosphere does not allow these radiations to pass through as such they heat up the earth’s atmosphere.

Question 3.
How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves.
Answer:
Electromagnetic waves are produced by accelerated charges which produce an oscillating electric field and magnetic field (which regenerate each other).

• Source of the Energy: Energy of the accelerated charge or the source that accelerates the charges.
• Expression: E_x = E_o sin (kz – ωt) and B_y = B_o sin (kz – ωt)
  (a) They are transverse in nature.
  (b) They don’t require a medium to propagate.

Question 4.
How are em waves produced by oscillating charges?
Draw a sketch of linearly polarised em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.
Answer:
(a) An oscillating charge produces an oscillating electric field in space, which produces an oscillating magnetic field. The oscillating electric and magnetic fields regenerate each other, and this results in the production of em waves in space.
(b) See Figure.

Question 5.
Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is i = ε_odϕEdt where ΦE is the electric flux produced during charging of the capacitor plates.
Answer:
The generalized form of Maxwell ampere law is
∮B⃗ ⋅dl→= μ_o(l + l_D) where l_D = ε_odϕEdt=dqdt

The electric flux Φ between the plates of the parallel plate capacitor through which a time-dependent current flow is given by:
Φ_E = E A, but E = σ/ε_o
Therefore we have

Question 6.
(a) Why are Infrared waves often called heatwaves? Explain.
Answer:
Infrared waves have frequencies lower than those of visibLe Light; they have the ability to vibrate not only the electrons but the entire atoms or molecules of a body. This vibration increases the internal energy and temperature of the body. That is why infrared waves are often called heat waves.

(b) What do you understand by the statement, “Electromagnetic waves transport momentum”? (CBSE AI, Delhi 2018)
Answer:
If we consider a plane perpendicular to the direction of propagation of the electromagnetic wave, then electric charges present on the plane will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges present on the surface thus acquire energy and momentum from the waves. This just illustrates the fact that an electromagnetic wave (like other waves) transfers energy and momentum.

Question 7.
(a) When the oscillating electric and magnetic fields are along the x- and y-direction respectively
(i) point out the direction of propagation of the electromagnetic wave,
Answer:
Z-axis

(ii) express the velocity of propagation in terms of the amplitudes of the oscillating electric and magnetic fields.
Answer:
c = E_o / B_o

(b) How do you show that the em wave carries energy and momentum? (CBSE A! 2013C)
Answer:
Consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If there are, on this plane, electric charges, they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. This illustrates the fact that an electromagnetic wave carries energy and momentum.

Question 8.
A radio can tune In to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band? (NCERT)
Answer:

Thus the wavelength band is 40 m to 25m.

Question 9.
Suppose that the electric field amplitude of an electromagnetic wave E₀ = 120 N C^-1 and that Its frequency is v = 50.0 MHz. (a) Determine, B0, ω k, and λ. (b) Find expressions for E and B. (NCERT)
Answer:


Question 10.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 V m^-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the 8 fields. [c = 3 × 10⁸ms^-1.]
Answer:

(c) The average density of electric field is
given by U_e = 12ε₀E² and the average energy density of the magnetic field is given by U_B = B22μ0. But B = Ec and C = 1μ0ε0√ , hence the above equation becomes UB = B22μ0=E22μ0c2,

In This Post we are  providing Chapter- 5 MAGNETISM AND MATTER NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON MAGNETISM AND MATTER

1. A particle of mass m and charge q moving with a uniform speed  normal to a uniform magnetic field B describes a circular path of radius & Derive expressions for (1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the particle?
Ans. A particle of mass (m) and change (q) moving with velocity normal to  describes a circular path if
 


Since Time period of Revolution
During circular path = 
=> ()
=> T = 

Kinetic energy K.E = 
=> KE = 

2. Write an expression for the force experienced by the charged particle moving in a uniform magnetic field B With the help of labeled diagram explain the working of cyclotron? Show that cyclotron frequency does not depend upon the speed of the particle?
Ans. Force experienced by the charged particle moving at right angles to uniform magnetic field  with velocity  is given by  = q () Initially Dee is negatively charged and Dee is positively charged so, the positive ion will get accelerated towards Dee since the magnetic field is uniform and acting at right angles to the plane of the Dees so the ion completes a circular path in when ions comes out into the gap, polarity of the Dee’s gets reversed used the ion is further accelerated towards Dee with greater speed and cover a bigger semicircular path. This process is separated time and again and the speed of the ion becomes faster till it reaches the periphery of the dees where it is brought out by means of a deflecting plate and is made to bombard the target.

Since F = qVBsin90⁰ provides the necessary centripetal force to the ion to cover a circular path so we can say 
=> r = 
Time period = 
V = 
 frequency is independent of velocity

3. (a) Obtain an expression for the torque acting on a current carrying circular loop.
(b) What is the maximum torque on a galvanometer coil 5 cm 12 cm of 600 turns when carrying a current of 10^-5 A. in a field where flux density is?
Ans. ABCD is a rectangular loop of length (L), breadth (b) and area (A). Let I be the Current flowing in the anti clockwise direction. Let  be the angle between the normal to the loop and magnetic field 

Force acting on arm AB of the loop

Force on arm CD

Force on arm BC

Force on arm DA

Since are equal and opposite and also acts along the same line, hence they cancel each other.
are also equal and opposite but their line of action is different, so they form a couple and makes the rectangular loop rotate anti clockwise.
Thus = either force  distance





For loop of N turns



Where M is magnet
ic moment of the loop.

Torque will be maximum when  = 90^o
 

4. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor of two. Calculate by what factor, the voltage sensitivity changes?
Ans. Current sensitivity 
Voltage sensitivity 
Resistance of a galvanometer increases when n and A are changed
Given  = 2R
Then n =  and A = 
New current sensitivity

New voltage sensitivity


From (i) and (iii)


n’A’=
Using equation (iv)



Thus voltage sensitivity decreases by a factor of .

5. (a) Show how a moving coil galvanometer can be converted into an ammeter?
(b) A galvanometer has a resistance 30 and gives a full scale deflection for a current of 2mA. How much resistance in what way must be connected to convert into?
(1) An ammeter of range 0.3A
(2) A voltammeter of range 0.2V.
Ans. (a) A galvanometer can be converted into an ammeter by connecting a low resistance called shunt parallel to the galvanometer.
Since G and R_S are in parallel voltage across then is same 

 
(b) (1) I = 0.3A G = 30 Ig = 2mA = 
Sheent (S) = 

S = 0.2
(2) G = 30, Ig = 2mA =, V = 0.2V
Shunt Resistance (R) 

R = 70 

6. 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 (me=). [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.]
Ans. Energy of an electron beam, E= 18 keV =
Charge on an electron, e= 
E=
Magnetic field, B = 0.04 G
Mass of an electron, me= 
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.

7. A sample of paramagnetic salt contains  atomic dipoles each of dipole moment . 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)
Ans. Number of atomic dipoles, n=
Dipole moment of each atomic dipole, M= 
When the magnetic field,  = 0.64 T
The sample is cooled to a temperature,  = 4.2°K
Total dipole moment of the atomic dipole, 
= 
= 
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, 
When the magnetic field,  = 0.98 T
Temperature,  = 2.8°K
Its total dipole moment =
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.

8. 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. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Ans. Number of horizontal wires in the telephone cable, n= 4
Current in each wire,  = 1.0 A
Earth’s magnetic field at a location, H= 0.39 G =
Angle of dip at the location, 
Angle of declination, 
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:
 Where,
B= Magnetic field at 4 cm due to current I in the four wires

 = Permeability of free space = 
= 0 = 0.2 G
∴ 
= 
The vertical component of earth’s magnetic field is given as:
Hv= Hsin

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


The resultant field at the point is given as:

s
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field:

= 0.39 cos 35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field:
 = 0.39

Angle, 
And resultant field:

9. 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.
Ans. 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.

10. 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 to roid 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 ferromagnetic 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?
Ans. (a)Owing to therandom thermal motion of 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) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater 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 and vice versa.
(e) The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.
(f) The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.

In This Post we are  providing Chapter- 6 ELECTROMAGNETIC INDUCTION NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROMAGNETIC INDUCTION

1. IF the rate of change of current of 2A/s induces an emf of 1OmV in a solenoid. What is the self-inductance of the solenoid?
Ans.

2. A circular copper disc. 10 cm in radius rotates at a speed of 2 rad/s about an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2T acts perpendicular to the disc.
1) Calculate the potential difference developed between the axis of the disc and the rim.
2) What is the induced current if the resistant of the disc is 2?
Ans. (1) Radius = 10cm, B = 0.2T w = 2 rad/s




I = 0.0314 A

3. An ideal inductor consumes no electric power in a.c. circuit. Explain?
Ans. P = E rms I rms cos 
But for an ideal inductor 

P=0

4. Capacitor blocks d.c. why?
Ans. The capacitive reactance

For d.c.  = 0

Since capacitor offers infinite resistance to the flow of d.c. so d.c. cannot pass through the capacitor.

5. Why is the emf zero, when maximum number of magnetic lines of force pass through the coil?
Ans. The magnetic flux will be maximum in the vertical position of the coil. But as the coil rotates 
Hence produced emf 

6. An inductor L of reactance  is connected in series with a bulb B to an a.c. source as shown in the figure.

Briefly explain how does the brightness of the bulb change when
(a) Number of turns of the inductor is reduced.
(b) A capacitor of reactance is included in series in the same circuit.
Ans. (a) Since Z = 
When number of turns of the inductor gets reduced and Z decreases and in turn current increases
Hence the bulb will grow more brightly
(b) When capacitor is included in the circuit

But (given)
Z = R (minimum)
Hence brightness of the bulb will become maximum.

7. 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  and the dip angle is 
Ans. Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 
Angle of dip, s
Vertical component of Earth’s magnetic field,
BV = B sin
= 
= 
Voltage difference between the ends of the wing can be calculated as:

=
= 3.125 V
Hence, the voltage difference developed between the ends of the wings is
3.125 V.

8. 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?
Ans. Mutual inductance of a pair of coils, µ = 1.5 H
Initial current,  = 0 A
Final current  = 20 A
Change in current, 
Time taken for the change, t = 0.5 s
Induced emf, 
Where is the change in the flux linkages 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.

9. A horizontal straight wire 10 m long extending from east to west is falling with a speed of , at right angles to the horizontal component of the earth’s magnetic field, Wb .
(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?
Ans. Length of the wire, l = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 
(a) Emf induced in the wire,
e = Blv


(b) Using Fleming’s right hand rule, its 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.

10. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad 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.
Ans. Length of the rod, l = 1 m
Angular frequency, = 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.
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.

In This Post we are  providing Chapter-1 ELECTRIC CHARGES AND FIELDS NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTRIC CHARGES AND FIELDS

Question 1.
Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point
(i) outside and
(ii) inside the shell.
Plot a graph showing variation of electric field as a function of r > R and r < R (r being the distance from the centre of the shell)
Answer:
(i) Field Outside Shell :
Consider a thin spherical shell of radius R with centre O. Let charge +q be distributed uniformly over the surface of shell. To calculate electric field intensity at P where OP = r, imagine a sphere S, with centre at O and radius r. The surface of sphere is Gaussian surface over at every point. Electric field is same and directed radially outwards.

Applying Gauss’ theorem

r→ is distance of point P from centre where E is calculated]
(ii) Inside Shell: As we know charge is located on its surface,

(iii) at r < R E→ is zero and r  = R, E is maximum at r > R, E is decreasing at E ∝ 1r2

Question 2.
Using Gauss’s law, derive the expression for the electric field at a point
(i) outside and
(ii) inside a uniformly charged thin spherical shell. Draw a graph showing electric field E as a function of distance from the centre.
Answer:
Electric field due to a uniformly charged spherical shell:
Suppose a thin spherical shell of radius R and centre O
Let the charge + q be distributed over the surface of sphere
Electric field intensity E→ is same at every point on the surface of sphere directed directly outwards
Let a point P be outside the shell with radius vector

Question 3.
(a) An electric dipole of dipole moment p→ consists of point charges + q and – q separated by a distance 2a apart. Deduce the expression for the electric field E→ due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment p→?. Hence show that in the limit x >> a, E→ —>2p→ (4πε₀x³).
(Delhi 2015)
Answer:
(a) Expression for magnetic field due to dipole on its axial lane:

(b)Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube given zero

Question 4.
(a) Define electric flux. Write its S.I. unit. “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is”. Justify this statement with the help of a suitable example.
(b) Use Gauss’s law to prove that the electric field inside a uniformly charged spherical shell is zero. (All India)
(a) Electric flux. The electric lines of force passing through that area, when held normally to the lines of force.
Answer:
Electric flux. The electric lines of force passing through that area, when held normally to the lines of force.

S.I. units: Vm, Nm²C^-1
Gauss’s Law states that the electric flux through a closed surface is given by

The law implies that the total electric flux through a closed surface depends on the quantity of total charge enclosed by the surface, and does not depend on its shape and size.

For example, net charge enclosed by the electric dipole (q, -q) is zero, hence the total elec¬tric flux enclosed by a surface containing electric dipole is zero.
(b) Electrical field inside a uniformly charged spherical shell. Let us consider a point ‘P’ inside the shell. The Gaussian surface is a sphere through P centred at O.

The flux through the Gaussian surface is E × 4πr².

However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives
E × 4πr² = 0
or E = 0
(r < R)
that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.

Question 5.
(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field.
(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor. (All India 2015)
Answer:
(a) (i) Energy of a parallel plate capacitor. Consider a capacitor of capacitance C. Initial charge on plates is zero. Initial potential difference between the capacitor plates is zero. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases. Let at any instant when charge on capacitor be q, the potential difference between its plates be,

Now work done in giving an additional infinitesimal charge dq to capacitor

The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works, which may be obtained by integration. Therefore total work


If V is the final potential difference between capacitor plates, then Q = CV

This work is stored as electrostatic potential energy of capacitor i.e., Electriostatic potential energy,

(ii) Expression for Energy Density.
Consider a parallel plate capacitor consisting of plates, each of area A, separated by a distance d. If space between the plates is filled with a medium of dielectric constant K, then
Capacitence of Capacitor,

If σ is the surface charge density of plates, then electric field strength between the plates.

This is the expression for electrostatic energy density in medium of dielectric constan K. In air of free space  (K = 1), therefore energy density,

(b) The energy of the capacitor when fully charged is

When this charged capacitor is connected to an identical capacitor C, then the charge will be distributed equally, q2 on each of the capacitors, then


Hence, the total energy stored is half of that stored initially in one capacitor which means the energy stored in combination is less than that stored initially in the single capacitor.

Question 6.
(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors Cj and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.
Answer:
(i)
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

(b) and

Question 7.
(a) Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the centre of the dipole on the axial line. (b) Draw a graph of E versus r for r >> a.
(c) If this dipole were kept in a uniform external electric field diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.
Answer:
(a) Expression for magnetic field due to dipole on its axial lane:

(b)Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube given zero

(b) Graph between E Vs r

(i) Diagrammatic representation
(ii) Torque acting on these cases
(i) In stable equilibrium, torque is zero (θ = 0)

(ii) In unstable equilibrium also, torque is zero (θ = 180°)

Question 8.
(a) Use Gauss’s theorem to find the electronic field due to a uniformly charged infinitely large plane thin sheet with surface charge density a.
(b) An infinitely large thin plane sheet has a uniform surface charge density +a. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet.

Answer:
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

Question 9.
(a) State Gauss’ law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density X.
(b) A wire AB of length L has linear charge density λ = kx, where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through this surface.

Answer:
(a)
Gauss’s law in electrostatics : It states that “the total electric flux over the surface S in vaccum is 1ε0 times the total charge (q).”

Electric field due to an infinitely long straight wire : Consider an infinitely long straight line charge having linear charge density X to determine its electric field at distance r. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface Sj and is directed radially outward.
Total flux through the cylindrical surface,

(b) Given : Length of wire = L, Charge density (λ) = kx, ϕ = ?
We know

Question 10.
Two point charges 4 (J.C and +1 pC are separated by a distance of 2 m in air. Find the point on the line-joining charges at which the net electric field of the system is zero. (Comptt. Outside Delhi 2017)
Answer:
q₁ = 4 µC, q₂ = 1 µC, r = 2 m
At this point, the net electric field of the system is zero.

In This Post we are  providing Chapter-2 ELECTROSTATIC POTENTIAL AND CAPACITANCE NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROSTATIC POTENTIAL AND CAPACITANCE

1. Explain what would happen if in the capacitor , a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
1. While the voltage supply remained connected.
2. After the supply was disconnected.
Ans. Dielectric constant of the mica sheet, k = 6
1. Initial capacitance, 
New capacitance, 
Supply voltage, V = 100 V
New charge s
Potential across the plates remains 100 V.
2. Dielectric constant, k = 6
Initial capacitance
New capacitance 
If supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge = 
Potential across the plates is given by,

2. A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Ans. Capacitor of the capacitance, 
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,















Therefore, the el
ectrostatic energy stored in the capacitor is 

3. 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?
Ans. 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  of the combination is given by,



New electrostatic energy can be calculated as



Loss in electrostatic energy =



Therefore, the electrostatic energy lost in the process is.

4. A spherical conducting shell of inner radius and outer radius has a charge
1. 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?
2. Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Ans. (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,
1. 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.

5. If one of the two electrons of a molecule is removed, we get a hydrogen molecular ion. In the ground state of an, the two protons are separated by roughly 1.5 , and the electron is roughly 1  from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Ans.

The system of two protons and one electron is represented in the given figure.
Charge on proton 1, 
Charge on proton 2, 
Charge on electron, 
Distance between protons 1 and 2, 
Distance between proton 1 and electron, 
Distance between proton 2 and electron, 
The potential energy at infinity is zero.
Potential energy of the system,

Substituting 



Therefore, the potential energy of the system is .

6. 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.
Ans .Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,


However 
And 

Putting the value of (2) in (1), we obtain

Therefore, the ratio of electric fields at the surface is.
 

7. 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  or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Ans. Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = Capacitance of a parallel plate capacitor is given by the relation,
Where,
 = Permittivity of free space = 

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of .

8.A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports

Show that the capacitance of a spherical capacitor is given by
where  and  are the radii of outer and inner spheres, respectively.
Ans.Radius of the outer shell = 
Radius of the inner shell = 
The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge  Potential difference between the two shells is given by,
Where,
= Permittivity of free space


Capacitance of the given system is given by,

=
Hence, proved.

9. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5. 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).
Ans.Length of a co-axial cylinder, l = 15 cm = 0.15 m
Radius of outer cylinder,  = 1.5 cm = 0.015 m
Radius of inner cylinder,  = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 
Capacitance of a co-axil cylinder of radii and is given by the relation,

Where,
 = Permittivity of free space = 


Potential difference of the inner cylinder is given by,

10. 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. (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?
Ans.Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material,  =3 Dielectric strength = 
For safety, the field intensity never exceeds 10% of the dielectric strength. Hence, electric field intensity, E = 10% of = 
Capacitance of the parallel plate capacitor, C = 50 pF =
Distance between the plates is given by,


Capacitance is given by the relation,

Where,
A = Area of each plate
= Permittivity of free space =


Hence, the area of each plate is about 19.Search

In This Post we are  providing Chapter- 3 CURRENT ELECTRICITY NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON CURRENT ELECTRICITY

Question 1.
A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current l.
It is found that when R = 4 Ω the current is 1 A and when R is increased to 9 Ω, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.
Answer:
The plots are as shown

Here l₁ = 1.0 A, R₁ =4 ohm, l₂ = 0.5 A, R₂ =9 ohm
Using the equation l = E(R+r) Or E = l(R + r)

we have
1.0 × (4 + r) = 0.5 × (9 + r)

Solving the above equation for r we have r = 1 ohm
Also E = 0.5 (9 + 1) = 5 V

Question 2.
A wire of resistance R, length l and area of cross-section A is cut into two parts, having their lengths in the ratio 1:2. The shorter wire is now stretched till its length becomes equal to that of the longer wire. If they are now connected in parallel, find the net resistance of the combination.
Answer:
Since the wires are cut in the ratio of 1:2 therefore,
Resistance of the shorter wire R₁ = R3 and

Resistance of the longer wire R₂ = 2R3

Since the shorter wire is stretched to make it equal to the longer wire therefore, it is stretched by n = 2 times its length. Hence New resistance of the shorter wire

Question 3.
In the figure, a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of EMFs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε₁/ε₂ and (ii) position of null point for the cell ε₁. How is the sensitivity of a potentiometer increased?

Answer:
From the diagram we have
ε1−ε2ε1+ε2=120300=25
Or
5ε₁ – 5ε₂ = 2ε₁ + 2ε₂

Solving we have
ε1ε2=73 ….(1)

Also let L be the balancing length for cell of emf ε1, then
ε1ε1+ε2=L300

Using equation (1) we have
ε1ε1+37ε1=L300

Solving for L we have L = 210 cm
The sensitivity of a potentiometer can be increased by increasing the length of the potentiometer wire.

Question 4.
The network PQRS, shown in the circuit diagram, has batteries of 4 V and 5 V and negligible internal resistance. A milli- ammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter.

Answer:
Using Kirchhoff’s junction rule to distribute current we have

Consider the loop SRPS, by Kirchhoff’s loop rule we have
200 l₂ + 20 (l₁ + l₂ ) – 5 = 0 …(1)
Or
220 l₂ + 20 l₁ = 5 …(2)

Consider the loop PRQP, by Kirchhoff’s loop rule we have
– 60 l₁ + 4 – 20 (l₁ + l₂) = 0 …(3)
80 l₁ + 20 l₂ =4 …(4)

Multiplying equation (2) by (4) we have
880 l₂ + 80 l₁ = 20 …(5)

Subtracting equation (4) from equation (5)
we have
860 l₂ = 16 or l₂ = 4/215 A

Substituting in equation (4) we have
l₁ = 39860A

Therefore reading of the milliammeter is
l₁+ l₂ = 4215+39860 = 0.063 A = 63 mA

Question 5.
A set of ‘n’ identical resistors, each of resistance ‘R’ when connected in series have an effective resistance ‘X’. When they are connected in parallel, their effective resistance becomes ‘Y’. Find out the product of X and Y.

Answer:
In series
R_s = R₁ + R₂ + R₃ + ……
R_s = X = R + R + R + …. upto n
X=nR

In Parallel

Question 6.
In the following circuit, a metre bridge is shown in its balanced state. The metre bridge wire has a resistance of 1 ohm per centimetre. Calculate the value of the unknown resistance X and the current drawn from the battery of negligible internal resistance.

Answer:
Using the Wheatstone bridge principle we have
4060=X3
or X = 2 Ω

Now total resistance of the combination is
R = 5×1005+100=500105 = 4.76 Ω

Current drawn is
l = V/R = 6/4.76 = 1.26 A

Question 7.
Calculate the electrical conductivity of the material of a conductor of length 3 m, area of cross-section 0.02 mm² having a resistance of 2 ohms.
Answer:
Given L = 3 m,
A = 0.02 mm2 = 0.02 × 10-6 m2
R = 2 ohm.

Using the equation
R = ρLA
Or
ρ = RAL

σ = LAR = 30.02×10−6×2 = 7.5 × 107 Sm-1

Question 8.
A potential difference of 2 volts is applied between points A and B has shown in the network drawn in the figure. Calculate (i) equivalent resistance of the network across the points A and B and (ii) the magnitudes of currents in the arms AFCEB and AFDEB.

Answer:
The circuit can be redrawn as shown below.

As seen the circuit is a balanced Wheatstone bridge; therefore the resistance in the arm CD is superfluous.
(i) Resistance of arm FCE = 2 + 2 = 4 Ω
Resistance of arm FDE = 2 + 2 = 4 Ω
Hence net resistance of the circuit between A and B is
R = 4×44+4=168 = 2 Ω

(ii) current in the arm AFCEB
l = V/R = 2/4 = 0.5 A

Current in the arm AFDEB
l = V/R = 2/4 = 0.5 A

Question 9.
A cell of emf E and internal resistance ‘r’ gives a current of 0.8 A with an external resistor of 24 ohms and a current of 0.5 A with an external resistor of 40 ohms.
Calculate
(i) emf E and
(ii) internal resistance ‘r’ of the cell.
Answer:
Given l₁ = 0.8 A, R₁ = 24 ohm l₂ = 0.5 A, R₂ = 40 ohm
Using the equation
E = l(R + r) we have
0.8 × (24 + r) = 0.5 × (40 + r)

Solving for r we have r = 2.67 ohm
Also E = 0.5( 40 + 2.67) = 21.3 V

Question 10.
In the circuit diagram of the metre bridge given below, the balance point is found to be at 40 cm from A. The resistance of X is unknown and Y is 10 ohms.
(i) Calculate the value of X;
(ii) if the positions of X and Y are interchanged in the bridge, find the position of the new balance point from A; and
(iii) if the galvanometer and the cell are interchanged at the balance point, would the galvanometer show any current.

Answer:
(i) Using the Wheatstone bridge principle we have
4060=X10
Or
X = 6.67 Ω

(ii) If X and Y are interchanged then
L(100−L)=106.67 solving for L we have
L = 59.9cm

(iii) The galvanometer will not show any current.