Two bodies of masses $m$ and $M$ ( $M >> m$ ) are attached to the two ends of a light string passing over a pulley (light and smooth). What is the tension in the string when the masses are in motion?

(M - m) g

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2 m g

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M g

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(m + M) g

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Solution

The correct option is B $2 m g$
Let $a$ be the acceleration of the system.
for mass $M$ , $M a = M g - T . . . . . (1)$
for mass $m$ , $m a = T - m g . . . . . (2)$
$(1) \times m, (2) \times M,$
$M m a = M m g - m T . . . . . (3)$
$M m a = M T - M m g . . . . . (4)$
$(3) - (4), 0 = 2 M m g - (M + m) T$
$T = \frac{2 M m g}{(M + m)} = \frac{2 M m g}{M} = 2 m g$ $(M + m \sim M$ as $M > m)$

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Two bodies of masses m and M(M>>m) are attached to the two ends of a light string passing over a pulley (light and smooth). What is the tension in the string when the masses are in motion?

The correct option is B 2mgLet a be the acceleration of the system.for mass M, Ma=Mg−T.....(1)for mass m, ma=T−mg.....(2)(1)×m,(2)×M, Mma=Mmg−mT.....(3)Mma=MT−Mmg.....(4)(3)−(4),0=2Mmg−(M+m)TT=2Mmg(M+m)=2MmgM=2mg (M+m∼M as M>m)

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