\[(a)\;1.50 \;J\quad (b)\;2.05\;J \quad (c)\; 4.25 \;J \quad (d)\;1.05\;J\]

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Tension in $1^{st}$ case $=T_1=\large\frac{mv_1^2}{R_1}$

$\qquad=\large\frac{0.8 \times (0.8)^2}{0.3}$$=1.71N$

Tension in II case$= T_2 =\large \frac{mv^2_2}{R_2}$

$\qquad=\large\frac{(0.8)(2.4)^2}{0.1}$$=46.1 \; N$

Work done = change in kinetic energy of the block in two cases

$\qquad\qquad=\large\frac{1}{2}$$m (v_2^2-v_1^2)$

$\qquad\qquad=\large\frac{1}{2} $$(0.8)[(2.4)^2-(0.8)^2]$

$\qquad\qquad=\large\frac{1}{2} $$\times (0.8) \times (5.76-0.64)$

$\qquad\qquad=2.05\;J$

Hence b is the correct answer.

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