question_answer

A circular disc of radius R is removed from a bigger circular disc of radius 2R, such that circumferences of the discs coincide. The centre of mass of the new disc is \[''\alpha R''\]from the centre of the bigger disc. The value of \[\alpha \] is:

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 \[\alpha =\frac{1}{3}\]