Question

In a vertical plane inside a smooth hollow thin tube a block of same mass as that of tube is released as shown in the figure. When it is slightly disturbed, it moves towards right. By the time the block reaches the right end of the tube, displacement of the tube will be (where 'R' is mean radius of tube): Assume that the tube remains in vertical plane.

• A. $\frac{2R}{\pi }$
• B. $\frac{4R}{\pi }$
• C. $\frac{R}{2}$
• D. R

Answer 1

The correct option is C $\frac{R}{2}$

Initially, the centre of mass of the whole system lies at a distance $R$ i.e. $X_{cm}=R$

As no external force is acting on the sysetm, thus COM of the system would lie at its initial position.

GIven : $x_1=R+x$ $x_2=x$ $m_1=m_2=m$ where $x$ is the displacement of the tube.

Using $X_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$\therefore$ $R=\frac{m(R+x)+mx}{m+m}$ $⟹x=-\frac{R}{2}$

Thus the tube will displace by $\frac{R}{2}$ towards left.