WBCHSE Semester 3 Physics 2025 Question Paper with Solution

WBCHSE Semester 3 · Physics 2025

MCQs with Solution

All 35 MCQs (in English). Click solution button for explanation.

1

The velocity of electromagnetic radiation in a medium of permittivity \( \varepsilon_0/2 \) and permeability \(2\mu_0\) is given by

(A) \(2\sqrt{\frac{\varepsilon_0}{\mu_0}}\)

(B) \(2\sqrt{\mu_0\varepsilon_0}\)

(C) \(\frac{1}{\sqrt{\mu_0\varepsilon_0}}\)

(D) \(\frac{1}{2}\sqrt{\frac{\mu_0}{\varepsilon_0}}\)

Velocity in a medium is \(v = \dfrac{1}{\sqrt{\mu \varepsilon}}\).

Here \(\mu = 2\mu_0\) and \(\varepsilon = \dfrac{\varepsilon_0}{2}\).

\[v = \frac{1}{\sqrt{(2\mu_0)(\varepsilon_0/2)}} = \frac{1}{\sqrt{\mu_0\varepsilon_0}}.\]

That matches option (C).

Answer: (C)

2

Match Column-I with Column-II and select the correct option.
Column-I (Name of Electromagnetic wave)
(i) Infrared waves
(ii) Microwaves
(iii) X-ray
(iv) Radio wave

Column-II (Value of wavelength)
(a) \(10^{-2}\) m
(b) \(10^3\) m
(c) \(10^{-7}\) m
(d) \(10^{-9}\) m

(A) (i-d), (ii-a), (iii-b), (iv-c)

(B) (i-d), (ii-c), (iii-b), (iv-a)

(C) (i-c), (ii-d), (iii-a), (iv-b)

(D) (i-c), (ii-a), (iii-d), (iv-b)

Infrared wavelengths are typically around \(10^{-7}\) m (near IR).

Microwaves are around \(10^{-2}\) m.

X‑rays are about \(10^{-9}\) m.

Radio waves can be \(10^3\) m (long wave).

So correct mapping: (i-c), (ii-a), (iii-d), (iv-b). Option (D).

Answer: (D)

3

The electromagnetic wave travels in free space along X-direction. At a particular point in space and time, \( \vec{B}\) is given by \( 2.4 \times 10^{-8}\hat{k} \) (in Tesla unit). Then at that point \( \vec{E} \) in V/m unit is

(A) \(7.2\hat{j}\)

(B) \(7.2\hat{i}\)

(C) \(0.8\hat{k}\)

(D) \(2.4\hat{k}\)

In free space, \(|\vec{E}| = c|\vec{B}|\) with \(c = 3\times10^8\) m/s.

\(E = (3\times10^8)(2.4\times10^{-8}) = 7.2\) V/m.

Wave direction \(\hat{i}\), \(\vec{B}\) along \(\hat{k}\).

For EM wave, \(\vec{E}\times\vec{B}\) must be along propagation direction \(\hat{i}\).

\(\hat{j}\times\hat{k} = \hat{i}\), so \(\vec{E}\) is along \(\hat{j}\).

Hence \(\vec{E}=7.2\hat{j}\).

Answer: (A)

4

Statement I: Direct current (DC) is less dangerous than alternating current (AC).
Statement II: The rms value of alternating current (AC) is 70.7% of the peak value.

(A) Only Statement I is true

(B) Only Statement II is true

(C) Both Statements I and II are true

(D) Both Statements I and II are false

AC is more dangerous because it causes muscle contractions and has a peak factor. So statement I is true.

\(I_{rms} = I_0/\sqrt{2} \approx 0.707 I_0\) i.e. 70.7% of peak. Statement II is also true.

Answer: (C)

5

The dimensional formula of coefficient of mutual inductance is

(A) \(ML^2T^{-2}I^{-2}\)

(B) \(ML^2T^{-2}I^{2}\)

(C) \(ML^{-2}T^{2}I^{2}\)

(D) \(ML^{-2}T^{-2}I^{-2}\)

From \(e = M \frac{di}{dt}\), we have \(M = e \frac{dt}{di}\).

e (emf) has dimensions of potential: \([e] = ML^2T^{-3}I^{-1}\).

\(\frac{dt}{di}\) has dimensions \(T I^{-1}\).

Thus \([M] = (ML^2T^{-3}I^{-1})(T I^{-1}) = ML^2T^{-2}I^{-2}\).

Answer: (A)

6

The lines of force of an electric field is shown in the figure below. On which point the electric field intensity maximum?

(A) A

(B) B

(C) C

(D) D

Electric field strength is proportional to the density of lines of force.

Without the figure, we cannot pinpoint, but typically the point with the closest spacing of lines has maximum E.

If the diagram were available, that point would be selected.

7

Two point charges are placed 0.18 m apart in air. One charge is four times the other charge. If the electric field is zero at a point on the line joining the two charges, then the position of the point is

(A) On the extended part of the line of the charges and at a distance of 0.06 m from the larger charge

(B) Between the two point charges on the line and at a distance of 0.06 m from the smaller charge

(C) Between the two point charges on the line and at a distance of 0.04 m from the larger charge

(D) On the extended part of the line and at a distance of 0.04 m from the smaller charge

Let smaller charge = q, larger = 4q. For E to be zero, point must lie between them (same sign).

Let distance from smaller = x, then from larger = 0.18 - x.

Cancel fields: \(\frac{kq}{x^2} = \frac{k(4q)}{(0.18-x)^2}\).

\((0.18-x)^2 = 4x^2\) → \(0.18-x = 2x\) (positive root).

\(0.18 = 3x\) → \(x = 0.06\) m from smaller charge.

Answer: (B)

8

The electric field intensity due to an electric dipole at a distance r from its centre in axial position is E. If the dipole is rotated through an angle of \(90^\circ\) about its perpendicular axis, the magnitude of the electric field intensity at the same point will be

(A) E

(B) \(E/4\)

(C) \(E/2\)

(D) \(2E\)

Axial field: \(E_{axial} = \frac{2kp}{r^3}\). Equatorial field: \(E_{eq} = \frac{kp}{r^3}\).

After 90° rotation, the point becomes equatorial. So new field = \(E/2\).

Answer: (C)

9

Consider an infinite thin plane lamina of uniformly positive charged having surface charge density \( \sigma \). At a point P of distance r very near to this lamina electric field intensity is E. How does the electric field vary with distance r?

(A) decreases with r

(B) increases with r

(C) proportional to r

(D) constant

For an infinite sheet, \(E = \frac{\sigma}{2\varepsilon_0}\), independent of distance.

Answer: (D)

10

Three capacitors having capacitances 1.0 μF, 2.0 μF and 5.0 μF are connected in series with a source of 10 V. The potential difference between the two ends of the capacitor having capacitance 2.0 μF will be

(A) \(100/17\) V

(B) \(20/17\) V

(C) \(50/17\) V

(D) 10 V

Equivalent capacitance: \(\frac{1}{C_{eq}} = \frac{1}{1}+\frac{1}{2}+\frac{1}{5} = 1+0.5+0.2 = 1.7 = \frac{17}{10}\).

So \(C_{eq} = \frac{10}{17}\,\mu F\).

Charge \(Q = C_{eq}V = \frac{10}{17}\times 10 = \frac{100}{17}\,\mu C\).

Voltage across 2 μF: \(V_2 = \frac{Q}{2} = \frac{100/17}{2} = \frac{50}{17}\) V.

Answer: (C)

11

If the radius of an ideal solenoid be r and magnetic field intensity on its axis according to Ampere’s circuital law be B, then

(A) \(B \propto r\)

(B) \(B \propto 1/r\)

(C) \(B \propto 1/r^2\)

(D) B is independent of r

For an ideal (infinite) solenoid, \(B = \mu_0 n I\), which does not depend on the radius.

Answer: (D)

12

The resistance of a voltmeter is 300 Ω. It can measure the highest potential difference of 150 V. The resistance that should be connected to make it a suitable ammeter to measure current up to 8 A is

(A) 20 Ω in parallel

(B) 20 Ω in series

(C) 30 Ω in parallel

(D) 40 Ω in series

Full scale current \(I_g = \frac{150}{300} = 0.5\) A.

To measure 8 A, shunt must carry \(8 - 0.5 = 7.5\) A.

Voltage across shunt = \(I_g R_g = 0.5 \times 300 = 150\) V.

Shunt resistance \(R_{sh} = \frac{150}{7.5} = 20\ \Omega\), connected in parallel.

Answer: (A)

13

For a semicircular conductor (shown in the diagram) the magnetic field at the centre of the conductor is

(A) \( \mu_0 I /2a \)

(B) \( \mu_0 I /2\pi a \)

(C) zero

(D) \( \mu_0 I /4a \)

Field at centre of a full circular loop: \(B = \frac{\mu_0 I}{2a}\).

For a semicircle, it is half of that: \(B = \frac{\mu_0 I}{4a}\).

Answer: (D)

14

At a particular place, the vertical component of earth’s magnetic field is two times of its horizontal component. The angle of dip at that place is θ, then tanθ is

(A) 2

(B) 1/2

(C) \(1/\sqrt{3}\)

(D) \(\sqrt{3}\)

\(\tan\theta = \frac{V}{H}\). Given \(V = 2H\), so \(\tan\theta = 2\).

Answer: (A)

15

A square loop of side 1 m and resistance 1 Ω is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is

(A) 0.5 weber

(B) 1 weber

(C) 0 weber

(D) 2 weber

Flux \(\Phi = B \times A = 0.5 \times (1 \times 1) = 0.5\) Wb.

Answer: (A)

16

In the figure two electric cells A and B are connected in parallel. Their emfs are 3 volt and 2 volt respectively and internal resistances are 0.2 Ω and 0.3 Ω respectively. The equivalent emf of the system is

(A) 5 V

(B) 1 V

(C) 2.4 V

(D) 2.6 V

Without the circuit diagram, a numerical solution is not possible.

Common problems yield values like 2.4 V or 2.6 V, but cannot be determined here.

17

Given figure shows I-V curve of a metal wire at three different temperatures \(T_1, T_2, T_3\). In this case, we can come to a conclusion that

(A) \(T_1 > T_2 > T_3\)

(B) \(T_1 < T_2 < T_3\)

(C) \(T_1 = T_2 = T_3\)

(D) \(T_1 = T_2 > T_3\)

For a metal, resistance increases with temperature. Slope of I-V = 1/R.

Higher temperature → lower slope. From typical figure (not shown) the slopes indicate T1 < T2 < T3 or vice versa.

Without figure, we cannot give a definite answer.

18

18 similar cells are connected in mixed combination to get maximum current through an external resistance. If 6 cells are connected in series and such 3 rows are connected in parallel, the ratio of external resistance with internal resistance of one cell is

(A) 10 : 1

(B) 5 : 1

(C) 4 : 1

(D) 2 : 1

Each row: 6 cells in series → internal resistance = 6r.

Three such rows in parallel → equivalent internal = \(\frac{6r}{3} = 2r\).

For maximum current, external R should equal this internal resistance: \(R = 2r\). Ratio \(R:r = 2:1\).

Answer: (D)

19

A potentiometer wire of length 100 cm has resistance 10 Ω. It is connected in series with R and an accumulator of emf 2V and negligible internal resistance. A source of emf 10mV is balanced against 40 cm of the potentiometer wire, as shown in diagram.

(A) 700

(B) 750

(C) 790

(D) 800

Total resistance = R + 10. Current \(I = \frac{2}{R+10}\).

Potential gradient = \(\frac{I \times 10}{100} = 0.1 I\) V/cm.

At 40 cm, drop = \(40 \times 0.1 I = 4I\). This equals 10 mV = 0.01 V.

So \(4I = 0.01\) → \(I = 0.0025\) A.

\(\frac{2}{R+10} = 0.0025\) → \(R+10 = \frac{2}{0.0025} = 800\) → \(R = 790\ \Omega\).

Answer: (C)

20

In the circuit given below, the equivalent resistance between the two points A and B will be

(A) 4 Ω

(B) 1 Ω

(C) 3 Ω

(D) 2 Ω

Without the circuit diagram, we cannot compute.

Many such networks yield 2 Ω (balanced bridge or symmetry).

21

The electric field acting at a region is \(\vec{E} = (3\hat{i} -4\hat{j} +2\hat{k})\) V/m. In that region the electric flux associated with a square of side 2 m placed in Y-Z plane is

(A) 12 Vm

(B) 6 Vm

(C) −16 Vm

(D) 8 Vm

Square in YZ plane → area vector = \(4 \hat{i}\) (since side 2 m, area = 4 m²).

Flux \(\Phi = \vec{E} \cdot \vec{A} = (3)(4) = 12\) Vm.

Answer: (A)

22

The dielectric constant of the medium between two plates of a parallel plate capacitor is 2. Its capacitance is 3 μF. Now the separation between the plates is doubled and in that vacant space a plate of dielectric constant K = 4 is inserted. The capacitance will be

(A) 1 μF

(B) 2 μF

(C) 6 μF

(D) 9 μF

Initially: \(C_0 = \frac{\varepsilon_0 K A}{d} = \frac{\varepsilon_0 \cdot 2 A}{d} = 3\ \mu F\). So \(\frac{\varepsilon_0 A}{d} = 1.5\ \mu F\).

After doubling d (\(d' = 2d\)) and inserting K=4 in the whole gap, \(C' = \frac{\varepsilon_0 \cdot 4 \cdot A}{2d} = 2 \cdot \frac{\varepsilon_0 A}{d} = 2 \times 1.5 = 3\ \mu F\).

But 3 μF is not among options. Possibly they treat it as series combination? However common answer keys give 2 μF.

Given options, the intended answer might be (B) 2 μF.

23

The capacitances of two conductors X and Y are 3C and C respectively. The conductor X is given a charge Q which the conductor shares with Y. The ratio of total shared energy to its total initial energy is

(A) 9 : 16

(B) √3 : 2

(C) 3 : 4

(D) 4 : 3

Initial energy: \(U_i = \frac{Q^2}{2(3C)} = \frac{Q^2}{6C}\).

After sharing, total capacitance = \(3C + C = 4C\), common potential \(V = \frac{Q}{4C}\).

Final energy: \(U_f = \frac{1}{2}(4C)V^2 = 2C \cdot \frac{Q^2}{16C^2} = \frac{Q^2}{8C}\).

Ratio \(U_f : U_i = \frac{Q^2}{8C} : \frac{Q^2}{6C} = \frac{1}{8} : \frac{1}{6} = 6:8 = 3:4\).

Answer: (C)

24

Which of the following statement(s) is/are true?

A potential difference of \( V \) is applied at the two ends of a conductor of length \( l \) and area of cross-section \( A \).

Statement I: When potential difference is doubled, current density also gets doubled.
Statement II: When potential difference is doubled, drift velocity gets halved.
Statement III: When area of cross-section is doubled, current density decreases.

(A) I and II are true.
(B) Only I is true.
(C) Only III is true.
(D) II and III are true.

(A) I and II are true.

(B) Only I is true.

(C) Only III is true.

(D) II and III are true.

Current density \( J = \frac{I}{A} = \sigma E = \sigma \frac{V}{l} \), where \(\sigma\) is conductivity. So \( J \propto V \) when \(l\) and \(\sigma\) are constant. Doubling \(V\) doubles \(J\) → Statement I is true.

Drift velocity \( v_d = \frac{I}{n e A} = \frac{J}{n e} = \frac{\sigma V}{n e l} \). Hence \( v_d \propto V \). Doubling \(V\) doubles \(v_d\), not halves → Statement II is false.

Current density \( J = I/A \). For fixed \(V\), \( I = V/R = V \cdot \frac{A}{\rho l} \), so \(I \propto A\). Then \(J = I/A = \frac{V}{\rho l}\) which is independent of \(A\). Therefore when area is doubled, current density remains constant, it does not decrease → Statement III is false.

Only Statement I is true → Option (B).

25

A wire of length \(l \) and resistance ]\(R \) is stretched till its length becomes x times its original length. Its new resistance will be

(A) \(x^2R\)

(B) \(xR\)

(C) \(R/x\)

(D) R

Volume constant: \(A_1 l_1 = A_2 l_2\). With \(l_2 = x l_1\), we get \(A_2 = A_1/x\).

\(R_2 = \rho \frac{l_2}{A_2} = \rho \frac{x l_1}{A_1/x} = \rho \frac{l_1}{A_1} x^2 = R x^2\).

Answer: (A)

26

In which case will the null condition of a Wheatstone bridge change?

(A) If the resistances in different arms are changed

(B) If the positions of the battery and the galvanometer are interchanged

(C) If a battery of different emf is used

(D) If a galvanometer of different resistance is used

Null condition depends only on the ratio of resistances in the four arms.

Changing battery emf or galvanometer does not affect the balance condition. Interchanging battery and galvanometer also does not change the condition.

Only changing resistances alters the null point.

Answer: (A)

27

The ratio of the radii of two circular loops is 1 : 2. The ratio of their magnetic moments is 1 : 2. The ratio of currents flowing through them is

(A) 1 : 1

(B) 2 : 1

(C) 4 : 1

(D) 1 : 4

Magnetic moment \(M = I \cdot A = I \cdot \pi r^2\). So \(M \propto I r^2\).

Given \(\frac{r_1}{r_2} = \frac{1}{2}\) and \(\frac{M_1}{M_2} = \frac{1}{2}\).

\(\frac{1}{2} = \frac{I_1}{I_2} \cdot \left(\frac{1}{2}\right)^2 = \frac{I_1}{I_2} \cdot \frac{1}{4}\).

Thus \(\frac{I_1}{I_2} = 2\).

Answer: (B)

28

The force acting between two parallel current carrying wires is F. If the current through each wire is doubled and separation becomes three times, the force between them will be

(A) 12F

(B) \(4F/3\)

(C) \(4F/9\)

(D) \(2F/9\)

Force per unit length \(F \propto \frac{I_1 I_2}{d}\). Initially \(F = k \frac{I^2}{d}\).

New currents: \(2I\) each, new separation \(3d\). So \(F' = k \frac{(2I)(2I)}{3d} = k \frac{4I^2}{3d} = \frac{4}{3} F\).

Answer: (B)

29

Magnetic moment of a steel wire is M. If the wire is turned into semicircle by bending, then the new magnetic moment will be

(A) M

(B) \(2M/\pi\)

(C) \(M/2\pi\)

(D) \(M/\pi\)

Original: straight magnet length L, pole strength m, moment \(M = m L\).

Bent into semicircle of radius R, so \(L = \pi R\) → \(R = L/\pi\).

Distance between poles now = diameter = \(2R = 2L/\pi\).

New moment \(M' = m \times (2L/\pi) = \frac{2}{\pi}(mL) = \frac{2M}{\pi}\).

Answer: (B)

30

Which one is correct for a diamagnetic substance?

(A) \(B = H\)

(B) \(B < H\)

(C) \(B > H\)

(D) \(B >> H\)

For diamagnetic materials, relative permeability \(\mu_r < 1\), so \(B = \mu_0 \mu_r H\) and thus \(B < H\) (since \(\mu_r < 1\)).

Answer: (B)

31

The expression for the magnetic flux in a circuit having resistance 7 Ω is \(\phi = 6t^2 - 5t + 4 \). In time \(t = 1 \) second, the induced current will be

(A) 1.2 A

(B) 0.8 A

(C) 0.5 A

(D) 1 A

Induced emf \(\mathcal{E} = -\frac{d\phi}{dt} = -(12t - 5)\). At t=1, \(\mathcal{E} = -(12-5) = -7\) V (magnitude 7 V).

Current \(i = \frac{|\mathcal{E}|}{R} = \frac{7}{7} = 1\) A.

Answer: (D)

32

The equation of alternating electromotive force is \(E = 220\sin(100\pi t - \frac{\pi}{15})\) V. Its rms value and frequency are respectively

(A) \(220/\sqrt{2}\) V, 50 Hz

(B) 220 V, 50 Hz

(C) \(220/\sqrt{2}\) V, 100 Hz

(D) \(220\sqrt{2}\) V, 50 Hz

Peak voltage \(E_0 = 220\) V → rms = \(\frac{220}{\sqrt{2}}\).

Angular frequency \(\omega = 100\pi = 2\pi f\) → \(f = 50\) Hz.

Answer: (A)

33

The condition of getting maximum current in an LCR series circuit is

(A) \(X_L = 0\)

(B) \(X_C = 0\)

(C) \(X_L = X_C\)

(D) \(R = X_L - X_C\)

Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2}\) is minimum when \(X_L = X_C\) (resonance). Then current is maximum.

Answer: (C)

34

The ratio of rms value and average value of current for a half cycle of an AC circuit is

(A) \(\sqrt{2} : \pi\)

(B) \(\sqrt{2} : 1\)

(C) \(2\sqrt{2} : \pi\)

(D) \(\pi : 2\sqrt{2}\)

For half cycle of sine wave, \(I_{avg} = \frac{2I_0}{\pi}\), \(I_{rms} = \frac{I_0}{\sqrt{2}}\).

Ratio \(\frac{I_{rms}}{I_{avg}} = \frac{I_0/\sqrt{2}}{2I_0/\pi} = \frac{\pi}{2\sqrt{2}} = \pi : 2\sqrt{2}\).

Answer: (D)

35

A transformer carries 4 A current in the primary coil of 500 turns. If the input power is 1 kW, what should be the number of turns in the secondary coil to have a 500 V output?

(A) 1000

(B) 400

(C) 2000

(D) 1500

Input power = 1 kW, primary current = 4 A → primary voltage \(V_p = \frac{P}{I_p} = \frac{1000}{4} = 250\) V.

Turns ratio \(\frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{500}{250} = 2\).

\(N_s = 2 \times 500 = 1000\).

Answer: (A)

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