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A sphere of radius 'a' and mass 'm' roll...

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed `v_(0)`. It encounters an inclined plane at angle `theta` and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel ?

A

`(2)/(5) (v_(0)^(2))/(gsintheta)`

B

`(v_(0)^(2))/(2g sin theta)`

C

`(v_(0)^(2))/(5g sin theta)`

D

`(7v_(0)^(2))/(10 g sin theta)`

Text Solution

Verified by Experts

The correct Answer is:
D

From energy conservation
`mgh=(1)/(2)mv_(0)^(2)+(1)/(2)Iomega^(2)`
`mgh=(1)/(2)mv_(0)^(2)+(1)/(2)xx(2)/(5)ma^(2)xx(v_(0)^(2))/(a^(2))`
`gh=(1)/(2)v_(0)^(2)+(1)/(5)v_(0)^(2)`
`gh=(7)/(10)v_(0)^(2)impliesh=(7v_(0)^(2))/(10g)`
From triangle `sintheta=(h)/(l)`
then `h=l sin theta`
`lsin theta=(7)/(10)(v_(0)^(2))/(g)`
`l=(7v_(0)^(2))/(10g sin theta)`
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