1.

In the above question, if the velocity of the object on reaching the bottom is upsilon for the smooth plane and(upsilon)/(n) for the rough plane, then the coefficient of friction is given by :

Answer»

`mu=tantheta[1-(1)/(n^(2))]`
`mu=costheta[1-(1)/(n^(2))]`
`mu=tantheta[1-(1)/(n^(2))]^((1)/(2))`
`mu=costheta[1-(1)/(n^(2))]^((1)/(2))`

Solution :Applying `upsilon^(2) - u^(2)= 2as` we have `upsilon^(2) = 2G sin thetaxxl`or `upsilon^(2) = 2g sin thetaxx L` and`(upsilon^(2))/(n^(2))=2g(sintheta-mucostheta)l` on
solving we GET the same value of .u. as in the above QUESTION (a) is the choice


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