A) \[2BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{1/2}}\]
B) \[BL\sin \left( \frac{\theta }{2} \right)(gL)\]
C) \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{3/2}}\]
D) \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{2}}\]
Correct Answer: A
Solution :
Þ \[h=L(1-\cos \theta )\] ??.(i) Maximum velocity at equilibrium is given by \ \[{{v}^{2}}=2gh=2g\,L(1-\cos \theta )\]\[=2g\ L\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)\] Þ \[v=2\sqrt{gL}\sin \frac{\theta }{2}\] Thus, max. potential difference \[{{V}_{\max }}=BvL\]\[=B\times 2\sqrt{gL}\sin \frac{\theta }{2}L\]\[=2BL\sin \frac{\theta }{2}{{(gL)}^{1/2}}.\]You need to login to perform this action.
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