Two masses $M_{1} = 5 k g$ and $M_{2} = 10 k g$ are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be:

g

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g / 2

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g / 3

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g / 4

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Solution

The correct option is C $g / 3$
Let the acceleration of the blocks and the tension in the string be $a$ and $T$ respectively.
For block of mass $M_{1}$ : $M_{1} a = T - M_{1} g$ .........(1)
For block of mass $M_{2}$ : $M_{2} a = M_{2} g - T$ .........(2)
Adding (1) and (2) we get $(M_{1} + M_{2}) a = (M_{2} - M_{1}) g$
$⟹$ $a = \frac{M_{2} - M_{1}}{M_{2} + M_{1}} g$
Given : $M_{1} = 5 k g$ $M_{2} = 10 k g$
$a = \frac{10 - 5}{10 + 5} g$ $⟹ a = \frac{g}{3}$

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Two masses M1=5kg and M2=10kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be:

Two masses $M_{1} = 5 k g$ and $M_{2} = 10 k g$ are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be: