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:-

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=1/2 g sin theta xx t_1^2` ….(i)
On rough inclined plane :
Acceleration , `a=(mg sin theta -muR)/m`
or `a=(mg sin theta -mu mg cos theta )/m`
`=g sin theta - mug cos theta`
`therefore s=1/2 (g sin theta - mug cos 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 , therefore n^2=(sin theta)/(sin theta - mu cos theta)`
or `mu=(n^2-1)/n^2 xx (sin theta)/(cos theta)` or `mu=(1-1/n^2) tan theta`