Question

A smooth ring P of mass $m$ can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley and carries a block Q of mass ($m/2$) as shown in the figure. At an instant, the string between the ring and the pulley makes an angle ${60}^o$ with the rod.
The initial acceleration of the ring is:

• A. $\frac{2g}{3}$
• B. $\frac{g}{6}$
• C. $\frac{2g}{9}$
• D. $\frac{g}{3}$

Answer 1

The correct option is B $\frac{g}{6}$

Consider a' be the initial acceleration of the ring

For ring : $Tcos{60}^o=m{a}^{\prime }=m\left(acos{60}^o\right)$

$⟹$ $T=ma$

For block : $\frac{m}{2}a=\frac{mg}{2}-T$

$\therefore$ $\frac{m}{2}a=\frac{mg}{2}-ma$ $⟹a=\frac{g}{3}$

Thus initial acceleration of the ring $a^{\prime }=acos{60}^o=\frac{g}{3}\times \frac{1}{2}=\frac{g}{6}$