In both the cases, block & monkey are at the same horizontal level. In both the cases, monkey climbs the rope. In case-1 rope remains vertical & in case-2 rope swings during motion. $t_{1} & t_{2}$ are times taken by monkeys to reach the pulley in case-1 & case-2 respectively. In both cases, monkey applies the same force on the rope.

t_{1} = t_{2}

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t_{1} < t_{2}

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In case-2, block reaches the pulley earlier than monkey

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In case-1, monkey reaches the pulley earlier than block

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Solution

The correct option is C In case-2, block reaches the pulley earlier than monkey
Monkey & block both reach the pulley at the same time. Upward acceleration
$= \frac{F - m g}{m}$
Time taken by monkey to reach pulley in case-2 is more as compared to case-1.
Block will reach in the same time, but monkey have less acceleration in upward direction, as only a component of ' $F$ ' helps monkey to reach pulley.
$t_{1} < t_{2}$ and in case-2, block reaches the pulley earlier than monkey.

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