The time taken by a body to slide down a rough `45^(@)` inclined plane is twice that required to slide down a smooth `45^(@)` inclined plane. The coefficient of kinetic friction between the object and rough plane is given by

Options

A. `sqrt((1)/(1 - n^(2))`
B. `sqrt(1 - (1)/(n^(2)))`
C. `1 - (1)/(n^(2))`
D. `(1)/(2 - n^(2))`

Answer

c

Explanation

From `s = at + (1)/(2)at^(2) = 0 + (1)/(2)at^(2) , r = sqrt((2s)/(a))`
For smooth plane ` a= g sin theta - mu cos theta`
For rough plane `a' = g (sin theta - mu cos theta)`
`:. t' = nt rArr sqrt((2s)/(g (in theta - mu cos theta)))= n sqrt((2s)/(gsin theta))`
`:.n^(2)g (sin theta - mu cos theta) = g sin theta `
where` theta = 45^(@) , sin theta = cos theta = 1//sqrt(2)`
Solving we get `mu = (1 - (1)/(n^2))`