Two masses M and M/2 are joined together by means of the light inextensible string passed over a frictionless pulley as shown in the figure.aWhen bigger mass is released the small one will ascend with an acceleration

g/3

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Solution

The correct option is A g/3
$F = M a$

$T - \frac{M g}{2} = \frac{M}{2} a$ ......(i)

$M g - T = m a$ ..........(ii)

Solving (i) and (ii)

$M g - \frac{M g}{2} = \frac{3 M}{2} a$

$a = \frac{g}{3}$

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Two masses M and M/2 are joined together by means of the light inextensible string passed over a frictionless pulley as shown in the figure.aWhen bigger mass is released the small one will ascend with an acceleration

The correct option is A g/3F=MaT−Mg2=M2a......(i)Mg−T=ma..........(ii)Solving (i) and (ii)Mg−Mg2=3M2aa=g3

The correct option is A g/3
$F = M a$

$T - \frac{M g}{2} = \frac{M}{2} a$ ......(i)

$M g - T = m a$ ..........(ii)

Solving (i) and (ii)

$M g - \frac{M g}{2} = \frac{3 M}{2} a$

$a = \frac{g}{3}$