Laws of Motion

JEE Main level, single correct, medium

27 of 210

Question 27 of 210

A monkey of mass 50kg50kg climbs on a rope which can withstand the tension (T)\left(T\right) of 350N350N. If monkey initially climbs down with an acceleration of 4ms−24m{s}^{-2} and then climbs up with an acceleration of 5ms−25m{s}^{-2}. Choose the correct option (g=10ms−2)\left(g=10m{s}^{-2}\right)

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Answer: (C) Rope will break while climbing upward

Let a1{a}_{1} be the acceleration of monkey while moving down.
F.B.D of monkey while moving downward


Using Newton's second law of motion,

mg−T=ma1mg-T=m{a}_{1}
∴500−T=50×4⇒T=300N∴500-T=50\times 4\Rightarrow T=300NLet a2{a}_{2} be the acceleration of monkey while moving up.
F.B.D of monkey while moving up


Using Newton's second law of motion

T−mg=ma2T-mg=m{a}_{2}
⇒T−500=50×5\Rightarrow T-500=50\times 5
⇒T=750N\Rightarrow T=750N
Breaking strength of rope
=750−300=350N=750-300=350N
∴∴ Rope will break while monkey is moving upward.

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