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A solid cylinder P rolls without slippin...

A solid cylinder P rolls without slipping from rest down an inclined plane attaining a speed vP at the bottom. Another smooth solid cylinder Q of same mass and dimensions slides without friction from rest down the inclined plane attaining a speed vQ at the bottom. The ratio of the speeds `((v_(Q))/(v_(P)))`is -

A

`(sqrt((3)/(4)))`

B

`(sqrt((3)/(2)))`

C

`(sqrt((2)/(3)))`

D

`(sqrt((4)/(3)))`

Text Solution

Verified by Experts

The correct Answer is:
B
If perfect rolling (solid cylinder P ) According to energy conservation law
`mgh= (1)/(2) mv_(p)^(2) + (1)/(2) I ((v_(p))/(R))^(2)`
`rArr mgh= (1)/(2) mv_(p)^(2) + (1)/(2) (mR^(2))/(2).(v_(p)^(2))/(R^(2)) rArr mgh= (3)/(4) mv_(p)^(2)`
`v_(p)^(2)= (4)/(3) gh`....(i)
if sliding without friction (solid cylinder Q ) According to energy conservation law
`mgh= (1)/(2) mv_(Q)^(2)`
`rArr v_(Q)^(2)= 2gh` ...(ii)
From (i) and (ii)
`(v_(Q)^(2))/(v_(P)^(2))= (2gh)/((4)/(3) gh)= (3)/(2) rArr (v_(Q))/(v_(p)) = sqrt(3//2)`
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