A) \[\sqrt{\frac{Bir}{m}}\]
B) \[\sqrt{\frac{Bir}{2m}}\]
C) \[\sqrt{\frac{2Bir}{m}}\]
D) \[2\sqrt{\frac{Bir}{m}}\]
Correct Answer: A
Solution :
[a] \[F\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,Bi\ell \] \[2T\,\sin \frac{\theta }{2}=F\] or \[2T\left( \frac{\theta }{2} \right)=Bi(r\theta )\] \[\therefore \,\,\,\,\,\,\,\,T=Bir\] Now speed, \[v=\sqrt{\frac{T}{m}}\] \[=\sqrt{\frac{Bir}{m}}\]You need to login to perform this action.
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