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A circular disc is rolling down an incli...

A circular disc is rolling down an inclined plane without slipping. If the angle of inclination is `30^(@)`, the acceleration of the disc down the inclined plane is :

A

g

B

`(g)/(2)`

C

`(g)/(3)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
`a=(gsintheta)/(1+(I)/(mr^(2)))`
For disc, `I=(1)/(2)mr^(2)impliesa=(gsin30^(@))/(1+(1)/(2)(mr^(2))/(mr^(2)))`
`a=(2)/(3)gxx(1)/(2)=(1)/(3)g`
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