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 -

Options

A. `(sqrt((3)/(4)))`
B. `(sqrt((3)/(2)))`
C. `(sqrt((2)/(3)))`
D. `(sqrt((4)/(3)))`

Answer

B

Explanation

If perfect rolling (solid cylinder P)
According to energy conservation law
`mgh=(1)/(2)mv_(p)^(2)+(1)/(2)I((v_(p))/(R))^(2)`
`impliesmgh=(1)/(2)mv_(p)^(2)+(1)/(2)(mR^(2))/(2).(v_(p)^(2))/(R^(2))impliesmgh=(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)`
`impliesv_(Q)^(2)=2gh`...........(ii)
From (i) and (ii)
`(v_(Q)^(2))/(v_(p)^(2))=(2gh)/((4)/(3)gh)=(3)/(2)implies(v_(Q))/(v_(P))=sqrt(3//2)`.