Q110 / 120 · JEE Main 2007 · Physics

A circular disc of radius $\mathrm{R}$ is removed from a bigger circular disc of radius $\mathrm{2R}$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\mathrm{\alpha / R}$ from the centre of the bigger disc. The value of $\mathrm{\alpha}$ is

  • Correct answer
    A
    $1 / 3$
  • B
    $1 / 2$
  • C
    $1 / 6$
  • D
    $1 / 4$

Solution

In this question distance of centre of mass of new disc is $\alpha R$ not $\frac{\alpha}{R}$. $\begin{aligned} & -\frac{3 \mathrm{M}}{4} \alpha \mathrm{R}+\frac{\mathrm{M}}{4} \mathrm{R}=0 \\ & \Rightarrow \alpha=\frac{1}{3} \end{aligned}$

Hence, the correct answer is option: 'A'.

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