An object of radius R and mass M is rolling horizontally without slipping with speed v . It then rolls up the hill to a maximum height `h = (3v^(2))/(4g)` . The moment of inertia of the object is ( g = acceleration due to gravity)

Options

A. `(2)/(5)MR^(2)`
B. `(MR^(2))/(2)`
C. `MR^(2)`
D. `(3)/(2)MR^(2)`

Answer

B

Explanation

Applying conservation of energy,
Initial KE = Final PE `rArr (1)/(2)Mv^(2)+(1)/(2)l omega^(2)=Mgh`
`(1)/(2)M+(1)/(2)(l)/(R^(2))=(3M)/(4)`
`rArr" "(1)/(2)M+(1)/(2)(l)/(R^(2))=(3M)/(4)`
`rArr" "(l)/(2R^(2))=((3)/(4)-(1)/(2))M`
`rArr" "l=2((1)/(4))MR^(2)=(1)/(2)MR^(2)`