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

Options

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)))`

Answer

A

Explanation

On smooth inclined plane:
Acceleration of the body `=g sin theta`
If s be the distance travelled, then
`s=1/2gsin thetaxxt_(1)^(2)`………I
On rough inclined plane:
Acceleration `a=(mg sin theta-muR)/m`
or `a=(mgsin theta-mu mg cos theta)/m`
`=g sin theta-mu g cos theta`
`:.s=1/29g sin theta-mu g vos theta t_(2)^(2)`.....ii
From equations i and ii
`(t_(2)^(2))/(t_(1)^(2))=(sin theta)/(sin theta-mu cos theta)`
But `t_(2)=nt_(1)," ":.n^(2)=(sin theta)/(sin thetas -mu cos theta)`
or `mu =(n^(2)-1)/(n^(2))xx(sin theta)/(cos theta)` or `mu=(1-1/(n^(2))) tan theta`