A) \[\frac{1}{4}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{6}\]
Correct Answer: B
Solution :
[b] O is centre of disc of radii 2R and mass M. \[{{C}_{1}}\] is centre of disc of radius R, which is removed. If \[\sigma \] is mass per unit area of disc. \[M=\pi {{(2R)}^{2}}\sigma \] Mass of disc removed \[{{M}_{1}}=\pi {{R}^{2}}\sigma \frac{M}{4}\] Mass of remaining disc \[{{M}_{2}}=M-\frac{M}{4}=\frac{3M}{4}\] \[{{M}_{1}}\times O{{C}_{1}}={{M}_{2}}\times O{{C}_{2}}\] \[\frac{M}{4}R=\frac{3M}{4}\times \Rightarrow x=\frac{R}{3}\] Compare with \[x=\alpha R\] We get \[\alpha =\frac{1}{3}\]You need to login to perform this action.
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