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

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 figure. When it is slightly disturbed it moves towards right. By the time the block reaches the right end of the tube then the displacement of the tube will be (where R is mean radius of tube). Assume that the tube remains in vertical plane and friction is abscent everywhere. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ku0porw4uhjyh19v.jpg" /> (A) (2 R/π ) (B) (4 R/π ) (C) (R/2) (D) R

Tardigrade Admin · Accepted Answer

Let the tube displaced by x towards left, then block will be displaced by (.R-x.) towards right ; <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ku0porw4uhjyr14j.jpg" /> mx=m(.R-x.)⇒ x=(R/2)

$\frac{2 R}{π}$

$\frac{4 R}{π}$

$\frac{R}{2}$

$R$

Let the tube displaced by $x$ towards left, then block will be displaced by $(R - x)$ towards right ;

$m x = m (R - x) \Rightarrow x = \frac{R}{2}$

π2R​

π4R​

2R​

R

Let the tube displaced by x towards left, then block will be displaced by (R−x) towards right ; mx=m(R−x)⇒x=2R​

$\frac{2 R}{π}$

$\frac{4 R}{π}$

$\frac{R}{2}$

$R$

Let the tube displaced by $x$ towards left, then block will be displaced by $(R - x)$ towards right ;

$m x = m (R - x) \Rightarrow x = \frac{R}{2}$