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A block moves down a smooth plane of inc...

A block moves down a smooth plane of inclination `theta`. Its velocity on reaching the bottom is v. If it slides down a rough inclined plane some inclination, its velocity on reaching the bottom is v/n, where n is a number greater than 0. The coefficient of friction is given by:

A

`mu = tan theta (1- 1/n^(2))`

B

`mu = cot theta (1-1/n^(2))`

C

`mu = tan theta (1- 1/n^(2))^(1//2)`

D

`mu = cot theta (1- 1/n^(2))^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`v^(2) = 2g sin theta xx l`
`v^(2)/n^(2) = 2(g sin theta - mu g costheta)l`
`sin theta(1-1/n^(2)) = mu cos theta`
`mu = tan theta (1-1/n^(2))`
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Knowledge Check

  • A block moves down a smooth inclined plane of inclination theta . Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of some inclination, its velocity on reaching the bottom is v/n, where n is a number greater than 0. The coefficient of friction is given by -

    A
    `mu=tantheta(1-(1)/(n^(2)))`
    B
    `mu=cottheta(1-(1)/(n^(2)))`
    C
    `mu=tantheta(1+(1)/(n^(2)))`
    D
    `mu=cottheta(1+(1)/(n^(2)))`

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    A
    `mu = (1-(1)/(n))`
    B
    `mu =(1-(1)/(n^2))`
    C
    `mu = sqrt(1-(1)/(n^2))`
    D
    `mu = sqrt(1-(1)/(n))`

  • An object is placed on the surface of a smooth inclined plane of inclination theta . It takes time t to reach the bottom. If the same object is allowed to slide down a rough inclined plane of inclination theta , it takes time nt to reach the bottom where n is a number greater than 1. The coefficient of friction mu is given by:-

    A
    `mu=tan theta(1-1/n^2)`
    B
    `mu= cot theta(1-1/n^2)`
    C
    `mu=tan theta sqrt(1-1/n^2)`
    D
    `mu=cot theta sqrt(1-1/n^2)`

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