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Question

In the system shown in the figure, blocks A and B have mass m1=2 kg and m2=267 kg respectively. A pulley having moment of inertia I=0.11 kgm2 can rotate without friction about a fixed axis. Inner and outer radii of the pulley are a=10 cm and b=15 cm respectively. B is hanging with the thread wrapped around the pulley, while A lies on a rough inclined plane . Coefficient of friction is μ=310 (g=10 ms2)


Column-I Column-II

i. Tension in the thread connecting block A
and the pulley (in N)
p. 17
ii. Tension in the thread connecting block B
and pulley (in N)
q. 26
iii. Angular acceleration of the pulley(in rads2) r. 20
iv. Acceleration of the block B (in ms2) s. 3

A
i-p ; ii-q ; iii-r ; iv-s
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B
i-q ; ii-r ; iii-s ; iv-p
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C
i-s ; ii-p ; iii-q ; iv-r
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D
i-p ; ii-r ; iii-q ; iv-s
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Solution

The correct option is A i-p ; ii-q ; iii-r ; iv-s
When the system is released, weight m2g tries to rotate the pulley in clockwise direction while the down-plane component m1gsin 30 of weight of block A tries to rotate it anticlockwise. But the moment produced by m2g is greater, therefore, the pulley has tendency to rotate clockwise.

Let its angular acceleration be α . Then acceleration of blocks A and B will be equal to aα (up the plane) and bα (vertically downwards) respectively.

Let tension in threads connected with blocks A and B be T1 and T2 respectively.
Considering free body diagrams,




For forces on block A,

N=m1gcos30
T1m1gsin30μN=m1(aα)

For forces on block B,
m2gT2=m2(bα)

Taking moments of forces acting on pulley, about axis of rotation,

T2bT1a=Iα

From the above equations,

N=103N
T1=17 N,T2=26 N,
α=20 rads2

Acceleration of block A=aα=2 ms2 (up the plane)
Acceleration of block B=bα=3 ms2 (downward)

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