A block of mass 2 kg hangs tangentially from the rim of a wheel of radius 0.5 m when released from rest the block falls vertically through 5 m height in 2 seconds the M.I. of the wheel is :

Options

A. `1kg.m^(2)`
B. `3.2kg.m^(2)`
C. `2.5kg.m^(2)`
D. `1.5kg.m^(2)`

Answer

D

Explanation

Here if h = height
Then `h=(1)/(2)at^(2)=(1)/(2)axx(2)^(2)`
`5=2aora=5//2`
Now the equation of motion is
`mg-T=ma`
`tau=T.R=Ialpha`
`=I.(a)/(R)`
or `I=(T.R^(2))/(a)`
`=I=((mg-ma)R^(2))/(a)`
`=(MR^(2)(g-a))/(a)`
`I=(2xx0.25[10-2.5])/(2.5)`
`=(7.5xx0.5)/(2.5)=1.5kgm^(2)`