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Question

# The pulley shown in figure has moment of inertia I=5 kg-m2 about its axis and radius R=1 m. Find the acceleration of the blocks if the masses of the blocks are M=3 kg and m=2 kg. Assume that the string is light and does not slip on the pulley and g=10 m/s2.

A
3 m/s2
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B
1 m/s2
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C
2 m/s2
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D
12 m/s2
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Solution

## The correct option is B 1 m/s2 From F.B.D of block M Mg−T1=Ma ... (i) From F.B.D of block m T2−mg=ma ... (ii) Torque equation about point (O) of pulley. T1R−T2R=Iα As string does not slip over the rim of the pulley, α=aR ⇒T1R−T2R=Iα=IaR ... (iii) Substituting T1 and T2 from equation (i) and (ii) [M(g−a)−m(g+a)]R=IaR Solving a=(M−m)gR2I+(M+m)R2 =(3−2)×10×125+(3+2)×12 =1010=1 m/s2

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