The pulley shown in the diagram is frictionless. A cat of mass 1 kg moves up on the massless string so as to just lift a block of mass 2 kg. After some time, the cat stops moving with respect to the string. The magnitude of the change in the cat’s acceleration is:

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\frac{g}{3}

\frac{2g}{3}

\frac{4g}{3}

answer is C.

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Detailed Solution

F.B.D of cat

$20 - 10 = a$

$a = 10 {ms}^{- 2}$

Question Image

$20 - T_{1} = 2 a \to (1)$

$T_{1} - 10 = a \to (2)$

$20 - T_{1} = 2 a$

$\underline{- 10 + T_{1} = a}$

$10 = 3 a$

$a = \frac{10}{3} {ms}^{- 2}$

$Δ a = 10 - \frac{10}{3}$

$= \frac{30 - 10}{3}$

$= \frac{20}{3} = \frac{2 \times 10}{3} = \frac{2g}{3}$

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