When a body slides down from rest along a smooth inclined plane making an angle of `30^(@)` with the horizontal, it takes time 20s. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it takes time 20p is, where p is some number greater than 1. The coefficient of friction between the body and the rough plane is

Options

A. `mu=(1-(1)/(p^(2)))(1)/(sqrt3)`
B. `mu=(1-(1)/(9p^(2)))`
C. `mu=(1-p^(2)) (1)/(sqrt3)`
D. `mu=sqrt(1-9p^(2))`

Answer

A

Explanation

The acceleration of the body sliding down smooth inclined plane is a=g `sin theta`
The acceleration of the same body sliding down a rough inclined plane is
`a'=g sing theta-mu g cos theta=g(sin theta-mu sin theta)....(ii) `
As the body slides a distance d in each case,
`therefore d=(1)/(2) at^(2)=(1)/(2)a't^(2) (therefore u=0)`
`therefore (a)/(a')=(t^('2))/(t^(2))=p^(2) (therefore t'=20p s, t=20s)`
`(gsintheta)/(g(sin theta-mu cos theta))=p^(2) ("Using (i) and (ii)")`
`sin theta=p^(2)sintheta-mup^(2)cos theta, mup^(2) cos theta=p^(2)sintheta-sintheta`
`mucos theta =((p^(2)-1))/(p^(2))sin theta, mu=tan theta (1-(1)/(p^(2))`
Here, `theta=30^(@) therefore mu, =((1-(1)/(p^(2))(1)/(sqrt(3))`