A) \[\frac{Ml}{M+m}\]
B) \[\frac{ml}{M+m}\]
C) \[\frac{M+m}{M}l\]
D) \[\frac{M+m}{m}l\]
Correct Answer: B
Solution :
If the both mass are revolving about the axis yy' and tension in both the threads are equal then \[M{{\omega }^{2}}x=m{{\omega }^{2}}(l-x)\] Þ \[Mx=m(l-x)\] Þ \[x=\frac{ml}{M+m}\]You need to login to perform this action.
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