Consider the situation shown in the figure. A | vedclass

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(C) Let $T$ be the tension in the string connected to the block of mass $m$. The tension in the string connected to the block of mass $2m$ will be $2T

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$\frac{2mg}{3}$

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(C) Let $T$ be the tension in the string connected to the block of mass $m$. The tension in the string connected to the block of mass $2m$ will be $2T$ because the movable pulley is supported by two segments of the same string.Let $a_A$ be the acceleration of the block of mass $2m$ and $a_B$ be the acceleration of the block of mass $m$.From the constraint relation,$a_B = 2a_A$.For the block of mass $m$ (hanging): $mg - T = m a_B = m(2a_A) \implies mg - T = 2ma_A \quad ...(1)$For the block of mass $2m$ (on the surface): $2T = (2m) a_A \implies T = ma_A \quad ...(2)$Substituting $(2)$ into $(1)$: $mg - ma_A = 2ma_A \implies mg = 3ma_A \implies a_A = \frac{g}{3}$.The tension in the string connected to $2m$ is $2T = 2(ma_A) = 2m(\frac{g}{3}) = \frac{2mg}{3}$.

Consider the situation shown in the figure. All the surfaces are smooth. The tension in the string connected to $2m$ is

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