Find the acceleration of the block of mass M in the situation shown in the figure. All the surfaces are frictionless and pulleys are massless.

g / 12

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2 g / 3

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g / 3

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g / 4

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Solution

The correct option is A $g / 3$
$T - M g s i n 30^{o} = M a$ .....(i)
$2 M g - 2 T = 2 M (\frac{a}{2})$ .....(ii)
Solving these equations, we get
$a = \frac{g}{3}$

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Find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light.

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Find the acceleration of the block of mass M in the situation shown in the figure. All the surfaces are frictionless and pulleys are massless.

The correct option is A g/3T−Mgsin30o=Ma .....(i)2Mg−2T=2M(a2) .....(ii)Solving these equations, we geta=g3

The correct option is A $g / 3$
$T - M g s i n 30^{o} = M a$ .....(i)
$2 M g - 2 T = 2 M (\frac{a}{2})$ .....(ii)
Solving these equations, we get
$a = \frac{g}{3}$