MHT CET · Physics · Oscillations

A simple pendulum with bob of mas ' \(m\) ' 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 ' \(\mathrm{B}\) '. Maximum e.m.f. induced across the pendulum is ( \(\mathrm{g}=\) acceleration due to gravity)

A \(\mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})\)
B \(2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^2\)
C \(2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^{3 / 2}\)
D \(2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^{1 / 2}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right)(\mathrm{gL})^{1 / 2}\)

Step-by-step Solution

Detailed explanation

We have, \(\mathrm{h}=\mathrm{L}(1-\cos \theta)\)
Maximum velocity at equilibrium position is given by,
\(\begin{aligned} & \mathrm{v}^2=2 \mathrm{gh}=2 \mathrm{gL}(1-\cos \theta)=2 \mathrm{gL}\left(2 \sin ^2\left(\frac{\theta}{2}\right)\right) \\ & \mathrm{v}=2 \sin \left(\frac{\theta}{2}\right) \sqrt{\mathrm{gL}}\end{aligned}\)
Thus Maximum potential difference,
\(\mathrm{V}_{\max }=\mathrm{BvL}=2 \mathrm{BL} \sin \left(\frac{\theta}{2}\right) \sqrt{\mathrm{gL}}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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