A) A
B) B
C) C
D) D
Correct Answer: B
Solution :
When string makes an angle \[\theta \] with the vertical in a vertical circle, then \[T-mg\cos \theta =\frac{m{{v}^{2}}}{l}\] or \[T=mg\cos \theta +\frac{m{{v}^{2}}}{l}\] Tension is maximum when cos \[\theta \] = + 1 i.e., \[\theta \] = 0. Thus, \[\theta \] is zero at lowest point B. At this point tension is maximum. So, string will break at point B. Note: The critical speed of a body n circular path \[{{v}_{c}}=\sqrt{Rg},\,R\]=radius of path. If at the highest point the speed is less than this the string would become slack and the body would leave the circular path.You need to login to perform this action.
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