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(D) Let $T$ be the tension in rope $A$. The painter pulls rope $A$ downwards with force $T$,and the rope pulls the painter upwards with force $T$. The

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Both $(A)$ and $(C)$.

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(D) Let $T$ be the tension in rope $A$. The painter pulls rope $A$ downwards with force $T$,and the rope pulls the painter upwards with force $T$. The platform is also supported by the other end of rope $A$,which exerts an upward force $T$ on the platform. Thus,the total upward force on the system (painter + platform) is $2T$.The rope $B$ supports the pulley,so the tension in rope $B$ is $T_B = 2T$. Given that the maximum tension rope $B$ can withstand is $300 \, N$,we have $2T_{\max} = 300 \, N$,which implies $T_{\max} = 150 \, N$.$(i)$ For maximum upward acceleration $a_{\max}$:Using Newton's Second Law for the system: $2T_{\max} - mg = m a_{\max}$.Given $mg = 200 \, N$,so $m = \frac{200}{g} \approx 20 \, kg$ (taking $g = 10 \, m/s^2$).$300 - 200 = 20 a_{\max} \implies 100 = 20 a_{\max} \implies a_{\max} = 5 \, m/s^2$. Thus,$(A)$ is correct.$(ii)$ When the painter is at rest,the system is in equilibrium:$2T = mg = 200 \, N \implies T = 100 \, N$. Thus,$(C)$ is correct.Since both $(A)$ and $(C)$ are correct,the correct option is $(D)$.

To paint the side of a building,a painter normally hoists himself up by pulling on the rope $A$ as shown in the figure. The painter and platform together weigh $200 \, N$. The rope $B$ can withstand a maximum tension of $300 \, N$. Then:

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