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° with the rod. The initial acceleration of the ring is

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° with the rod. The initial acceleration of the ring is (A) (2 g/3) (B) (2 g/6) (C) (2 g/9) (D) ( g /3)

Tardigrade Admin · Accepted Answer

(m/2) g -T=(m/2) a....(i) T cos 60°=( ma / cos 60°) ....(ii) Solving (i) and (ii) acceleration of ring =(2 g /9)

$\frac{2 g}{3}$

$\frac{2 g}{6}$

$\frac{2 g}{9}$

$\frac{g}{3}$

$\frac{m}{2} g - T = \frac{m}{2} a ....$ (i)
$T cos 6 0^{\circ} = \frac{ma}{c o s 6 0 ^{\circ}} ....$ (ii)
Solving (i) and (ii)
acceleration of ring $= \frac{2 g}{9}$

32g​

62g​

92g​

3g​

2m​g−T=2m​a....(i) Tcos60∘=cos60∘ma​....(ii) Solving (i) and (ii) acceleration of ring =92g​

$\frac{2 g}{3}$

$\frac{2 g}{6}$

$\frac{2 g}{9}$

$\frac{g}{3}$

$\frac{m}{2} g - T = \frac{m}{2} a ....$ (i)
$T cos 6 0^{\circ} = \frac{ma}{c o s 6 0 ^{\circ}} ....$ (ii)
Solving (i) and (ii)
acceleration of ring $= \frac{2 g}{9}$