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A small pulley of radius 20 cm and momen...


A small pulley of radius 20 cm and moment of inertia `0.32kg-m^(2)` is used to hang a 2 kg mass with the help of massless string. If the block is released, for no slipping condition acceleration of the block will be

A

`2m//s^(2)`

B

`4m//s^(2)`

C

`1m//s^(2)`

D

`3m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A


`20-T=2a` ..(i)
For pulley
`TR=Ialpha`
`thereforeT(0.2)=(0.32)alpha`
`T=1.6alpha` ..(ii)
`a=Ralpha=0.2alpha` ..(iii)
solving these equation we get
`a=2m//s^(2)`
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