A small sphere of radius R is held against the inner surface of larger sphere of radius `6R` (as shown in figure). The masses of large and small spheres are `4M` and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the large spheres, when the smaller sphere reaches the other extreme position.

Answer

B

Explanation

Since, all the surfaces are smooth, no external force is acting on the system in horizontal direction. Therefore, the centre of mass of the system in horizontal direction remains stationary.

x-coordinate of COM initially will given by
`x_i=(m_1x_1+m_2x_2)/(m_1+m_2)`
`x_i=((4M)(L)+M(L+5R))/(4M+M)=(L+R)` ...(i)
Let `(x,0)` be the coordinates of the centre of large shpere in final position. Then, x-coordinate of COM finally will be
`x_f=((4M)(x)+M(x-5R))/(4M+M)=(x-R)` ...(ii)
Equating Eqs. (i) and (ii), we have
`x=L+2R`
Therefore, coordinates of large sphere, when the smaller sphere reaches the other extreme position are `(L+2R,0)`.