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JEE Main 2022 · 26 Jul · Shift 1, Physics question 31

Three masses M = 100 kg, m1m_1 = 10 kg and m2m_2 = 20 kg are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m2m_2 moves upward with an acceleration of 2 ms−2^{-2}. The value of F is :

(Take g=10g=10 ms−2^{-2})

Block M = 100 kg on a frictionless floor pushed to the right by F; m1 = 10 kg rests on top of M and is tied by a string over a pulley at M's top right corner to m2 = 20 kg hanging against M's right face.
Choose one option
Answer and worked solution

Official answer: C

Hanging mass. m2m_2 rises at 2 m s−2^{-2}: T=20(10+2)=240T=20(10+2)=240 N.

Whole system. For m2m_2 to rise with m1m_1 pulled by the same string, the frame M (with m1m_1 and m2m_2) accelerates at a=Tm1=24a=\frac T{m_1}=24 m s−2^{-2} in NTA's intended reading, so F=(100+10+20)×24=3120F=(100+10+20)\times24=3120 N.

(C)

Topics:Constraint motion and pulleysPseudo forces and non-inertial frames
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