When body slides down from rest along smooth inclined plane making angle of `45^(@)` with the horizontal, it takes time `T` When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance it is seen to take time `pT`, where p is some number greater that 1. Calculate late the coefficient of friction beween the body and the rough plane.
.

Options

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

Answer

B

Explanation

Given situations are shown in figure, For smooth inclined plane, Here, `Msin theta=Ma`
`Rightarrow a=sin theta`
Let s=Length of inclined plane
`"Using", s=mt+(1)/(2)at^(2) Rightarrow s=0 xx T +(1)/(2)(g sin theta)T^(2) (therefore t=T)`
`Rightarrow s=(1)/(2)g sin theta T^(2) Rightarrow s=(1)/(2sqrt(2))g T^(2) (therefore theta =45^(@))...(i) `
For rough inclined plane.
`f=muN=muNcos theta`
`Mgsin theta-f=Ma'`
`Rightarrow a'=(sin theta-mucos theta)g`
`"Using" s=ut+(1)/(2)a't^(2)`
`Rightarrow s=0 xx(pT)+(1)/(2)(sin theta-mucos theta)g xx p^(2)T^(2)(therefore t=pT)`
`Rightarrow s=(1)/(2sqrt2)(1-mu)gp^(2)T^(2)...(ii)`
From equation (i) and (ii) `(1)/(2sqrt2)gT^(2)=(1)/(2sqrt2)(1-mu)gp^(2)T^(2)`
`Rightarrow 1=(1-mu)p^(2) Rightarrow mu=(1-(1)/(2))`