A) 1/3
B) 1/2
C) 1/6
D) 1/4
Correct Answer: A
Solution :
The distance of centre of mass of new disc from the centre of mass of remaining disc is \[\alpha R.\] Mass of remaining disc \[=M-\frac{M}{4}=\frac{3M}{4}\] \[\therefore \] \[-\frac{3M}{4}\alpha R+\frac{M}{4}R=0\] \[\Rightarrow \] \[\alpha =\frac{1}{3}\]You need to login to perform this action.
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