MHT CET 2025 20th April Evening Shift | Electromagnetic Induction Question 13 | Physics

A simple pendulum with bob of mass m and conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of earth's magnetic field in the direction perpendicular to swing is B . Maximum e.m.f. induced across the pendulum is ( $\mathrm{g}=$ acceleration due to gravity)

$2 \mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin \frac{\theta}{2}\right)$

$\mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin \frac{\theta}{2}\right)$

$\mathrm{BL}(\sqrt{\mathrm{gL}})^2\left(\sin \frac{\theta}{2}\right)$

$2 \mathrm{BL}(\sqrt{\mathrm{gL}})\left(\sin ^2 \frac{\theta}{2}\right)$

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

Three inductances are connected as shown in the figure. The equivalent inductance between A and b is

MHT CET 2025 20th April Evening Shift Physics - Electromagnetic Induction Question 13 English

2.25 H

1.20 H

0.225 H

0.120 H

MHT CET 2025 20th April Morning Shift

MCQ (Single Correct Answer)

-0

The current flowing through an inductor of selfinductance L is continuously increasing at constant rate. The variation of induced e.m.f. (e) verses $\mathrm{dI} / \mathrm{dt}$ is shown graphically by figure

MHT CET 2025 20th April Morning Shift

MCQ (Single Correct Answer)

-0

The coefficient of mutual induction is 2 H and induced e.m.f. across secondary is 2 kV . Current in the primary is reduced from 6 A to 3 A . The time required for the change of current is

$4 \times 10^{-3} \mathrm{~s}$

$6 \times 10^{-3} \mathrm{~s}$

$2 \times 10^{-3} \mathrm{~s}$

$3 \times 10^{-3} \mathrm{~s}$

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