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 :
\[mr{{\omega }^{2}}=MR{{\omega }^{2}}\] Put \[r=l-R\] \[\therefore \] \[m(l-R)=MR\] \[\Rightarrow \] \[R=\frac{ml}{M+m}\]You need to login to perform this action.
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