A body takes time `t` to reach the bottom of a smooth inclined plane of angle `theta` with the horizontal. If the plane is made rough, time taken now is `2t`.The coefficient of friction of the rough surface is

Options

A. `(3)/(4)tan theta`
B. `(1)/(2)tan theta`
C. `(2)/(3)tan theta`
D. `(1)/(4)tan theta`

Answer

a

Explanation

When the plane is smooth
`a = g sin theta, t_(1) = sqrt((2s)/(a)) = sqrt((2s)/g sin theta`
When plane is rough `a = g (sin theta - mu cos theta)`
`:. t_(2) = sqrt((2s)/(g(sin theta-mu cos theta)))`
As `t_(2) =2t_(1)`
`:. sqrt((2s)/(g (sin theta -mu cos theta))) = 2 sqrt((2s)/(g sin theta)`
Squaring we get
`(1)/(sin theta - mu cos theta)=(4)/(sin theta)`
`sin theta = 4 sin theta - 4 mu cos theta`
`4 mu cos theta =3 sin theta`
or `mu = (3)/(4)tan theta`.