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A cylinder rolls down an inclined plane ...

A cylinder rolls down an inclined plane of inclination `30^@`, the acceleration of cylinder is

A

`(g)/(2)`

B

`(g)/(3)`

C

g

D

`(2)/(3)g`

Text Solution

Verified by Experts

The correct Answer is:
B
Acceleration of a body rolling without slipping down an inclined plane is
`a=(gsin theta)/(1+(K^(2))/(R^(2)))`
Here , `theta=30^(@)`
For solid cylinder, `K^(2)=(1)/(2)R^(2)`
:. `a=(g sin 30^(@))/(1+(1)/(2))=(2)/(3)g sin 30^(@)=(g)/(3)`
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