Question

A small sphere of radius $R$, is held against the inner surface of a smooth spherical shell of radius $6R$ as shown in the figure. The masses of the shell and small sphere are $4M$ and $M$ respectively. This arrangement is placed on a smooth horizontal table. The small sphere is now released. The x-coordinate of the centre of the shell when the smaller sphere reach the other extreme position is,

• A. $R$
• B. $2R$
• C. $3R$
• D. $4R$

Answer 1

The correct option is B $2R$

Initially, x-coordinate of centre of mass is,
$x_i=\frac{\left(4M\right)\left(0\right)+M\left(5R\right)}{4M+M}=R-\left(i\right)$

Let $x_0$ be the x-coordinate of centre of the shell, when the small sphere reaches the other extreme position. Then the final x-coordinate of the centre of mass is,

$x_f=\frac{\left(4M\right)\left(x_0\right)+M(x_0-5R)}{4M+M}$
$=\frac{5M{x}_0-5MR}{5M}=x_0-R-\left(ii\right)$

As, all the surfaces are smooth, therefore, centre of mass will not move in x-direction.

$\therefore x_i=x_f$
$\Rightarrow R=x_0-R$
$\therefore x_0=2R$

Hence, $\left(B\right)$ is the correct answer.