A trolley T of mass 5 kg on a horizontal smooth surface is pulled by a load of 2 kg through a uniform rope ABC of length 2 m and mass 1 kg. As the load falls from BC = 0 to BC = 2 m, its acceleration (in ms^-2) changes from

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$\frac{20}{8} to \frac{30}{8}$

$\frac{20}{6} to \frac{30}{6}$

$\frac{20}{5} to \frac{30}{6}$

None of these

answer is A.

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Detailed Solution

$Initial force = 2 g = 20 N$

$Initial acceleration = \frac{Force}{Mass} = \frac{20}{5 + 1} = \frac{20}{6} {ms}^{- 2}$

$Final force = (Load + Mass of thread) \times g$

$= (2 + 1) \times 10 = 30 N$

$Therefore, final acceleration = \frac{30}{6} {ms}^{- 2}$

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