Explanation

(c) In the given condition tension in the string
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$T = \frac{2m_{1}m_{2}}{m_{1} + m_{2}}g = \frac{2 \times 0.36 \times 0.72}{1.08} \times 10$
$T = 4.8\ N$
And acceleration of each block
$a = \left( \frac{m_{2} - m_{1}}{m_{1} + m_{2}} \right)g = \left( \frac{0.72 - 0.36}{0.72 + 0.36} \right)g = \frac{10}{3}\ m/s^{2}$
Let ‘S’ is the distance covered by block of mass $0.36\ kg$ in first sec
$S = ut + \frac{1}{2}\ at^{2} \Rightarrow S = 0 + \frac{1}{2}\left( \frac{10}{3} \right) \times 1^{2} = \frac{10}{6}\ meter$
$\therefore$ Work done by the string $W = TS = 4.8 \times \frac{10}{6}$
$\Rightarrow W = 8\ Joule$