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(A) Initially,the system is in equilibrium. Let $T$ be the tension in the string. For the disc of mass $4m$,the tension in the two supporting strings

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$4$

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(A) Initially,the system is in equilibrium. Let $T$ be the tension in the string. For the disc of mass $4m$,the tension in the two supporting strings is $2T$. Thus,$2T = 4mg \implies T = 2mg$.When string $3$ is burnt,the block of mass $m$ is no longer held. The tension $T$ acts on the block. The frictional force on the block is $f = \mu mg = 0.5 	imes m 	imes 10 = 5m$.The equation of motion for the block is $T - f = ma \implies 2mg - 5m = ma \implies 20m - 5m = ma \implies a = 15 \ m/s^2$.However,the disc is supported by a movable pulley system. Let the acceleration of the disc be $a_d$. From the constraint relation,the acceleration of the block $a_b$ is related to $a_d$. Since the string length is constant,$a_b = 2a_d$.Using the equations of motion: For the block,$T - f = m(2a_d)$. For the disc,$4mg - 2T = 4m(a_d)$.Adding the equations: $4mg - f = m(2a_d) + 4m(a_d) = 6ma_d$.$4m(10) - 0.5m(10) = 6ma_d \implies 40m - 5m = 6ma_d \implies 35m = 6ma_d \implies a_d = 35/6 \approx 5.83 \ m/s^2$.Re-evaluating the setup: The tension $T$ is constant throughout the string. For the disc,$4mg - 2T = 4ma_d$. For the block,$T - f = ma_b$. With $a_b = 2a_d$,$T - 5m = m(2a_d)$.Substituting $T = 2a_d m + 5m$ into the disc equation: $40m - 2(2a_d m + 5m) = 4ma_d \implies 40m - 4a_d m - 10m = 4ma_d \implies 30m = 8ma_d \implies a_d = 30/8 = 3.75 \ m/s^2$. Given the options,the closest integer value is $4 \ m/s^2$.

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