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A uniform solid cylinder of mass 'm' can rotate freely about its axis which is kept horizontal. A particle of mass m0 hangs from the end of a light string wound round the cylinder. When the system is allowed to move, the acceleration with which the particle descends is


A
2m0gm0+2m
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B
2m0gm+2m0
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C
m0gm+m0
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D
2m0gm+2m0
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Solution

The correct option is B 2m0gm+2m0

Suppose that the tension in the string is T.
Then, =m0gT=m0a
T=m0(ga)
where a = acceleration
Further, T.r = I α
α = angular acceleration and T.r = moment of force acting on cylinder
or T=Iαr
=(mr22)(αr)=mrα2=ma2
a=2Tm=2m0(ga)m
Solving for a, we get
a=(2m0gm+2m0).


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