Question

A light smooth inextensible string connected with two identical particles each of mass $m$ passes through a light ring $R$. If a force $F$ pulls the ring, the relative acceleration between the particles has magnitude

• A. Zero
• B. $\frac{F}{m}\tan \theta$
• C. $\frac{F\sec \theta }{m}$
• D. $\frac{2F}{m}\tan \theta$

Answer 1

The correct option is B $\frac{F}{m}\tan \theta$

Let tension $T$ be induced in the string when force $F$ is applied on the ring.


From free body diagram, we have
$F=2T\cos \theta$

Let, the acceleration of each particle along the string is $a$, so from Newton's second law of motion,

$\Rightarrow T=\frac{F}{2\cos \theta }=ma$

$\Rightarrow a=\frac{F}{2m\cos \theta }$

The relative acceleration in $\text{x-}$direction is zero. There is a relative acceleration only in $\text{y-}$direction.

So, relative acceleration along $\text{y-}$direction,
$a_r=a_y-(-a_y)=2ay$

From figure, $a_r=2a\sin \theta [\because a_y=a\sin \theta ]$

Substituting the value of $a$, $a_r=2\frac{F\sin \theta }{2m\cos \theta }$

$\Rightarrow a_r=\frac{F}{m}\tan \theta$.

Hence, option (b) is the correct answer.