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A block of mass 2kg hangs from the rim o...

A block of mass 2kg hangs from the rim of a wheel of radius 0.5m. On releasing from rest the block falls through 5m height in 2s. The moment of inertia of the wheel will be

A

`1 kg-m^(2)`

B

`3.2 kg-m^(2)`

C

`2.5kg-m^(2)`

D

`1.5kg-m^(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
On releasing from rest the block falls through 5m height in 2 sec
`5= 0 +(1)/(2) a(2)^(2)` [As `S= u t+ (1)/(2) a t^(2)`]
`therefore a= 2.5m//s^(2)`
Substituting the value of a in the formula `a= (g)/(1 + (I)/(mR^(2)))` and by sloving we get
`rArr 2.5= (10)/(1+ (I)/(2 xx (0.5)^(2))) rArr I= 1.5kg -m^(2)`
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