Question 84 of 210

All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass $2kg$ is:

A$g$B$\dfrac{g}{3}$C$\dfrac{g}{2}$D$\dfrac{g}{4}$

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Answer: (B) $\dfrac{g}{3}$

For $4kg$ block:
$40-2T=4a \Rightarrow T=20-2a$
For $2kg$ block:
$T-10=4a \Rightarrow 20-2a-10=4a \Rightarrow 10=6a \Rightarrow a=\dfrac{10}{6}=\dfrac{g}{6}m{s}^{-2}$
Acceleration of the block of mass $2kg$, $2a=\dfrac{g}{3}m{s}^{-2}$.

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