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) \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{2}}\]
E) \[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{5/2}}\]
Correct Answer: A
Solution :
\[h=L-L\cos \theta \] \[\Rightarrow \] \[h=L(1-\cos \theta )\] ...(1) \[\therefore \] \[{{v}^{2}}=2gh=2gL(1-\cos \theta )\] \[=2g\,L\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)\] \[\Rightarrow \] \[v=2\sqrt{gL}\sin \frac{\theta }{2}\] Thus, maximum potential difference \[{{V}_{\max }}=BvL\] \[=B\times 2\sqrt{gL}\sin \frac{\theta }{2}L\] \[=2BL\sin \frac{\theta }{L}{{(gL)}^{1/2}}\]You need to login to perform this action.
You will be redirected in
3 sec