# A body takes time t to reach the bottom of a smooth inclined plane of angle θ with the horizontal…..

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

(a) $\displaystyle \frac{3}{4} tan\theta$

(b) $\displaystyle \frac{2}{3} tan\theta$

(c) $\displaystyle \frac{1}{4} tan\theta$

(d) $\displaystyle \frac{1}{2} tan\theta$

Ans: (a)

Sol: For Smooth surface

$\displaystyle S = \frac{1}{2}a t^2$ …(i)

For Rough surface

$\displaystyle S = \frac{1}{2}a’ (2t)^2$ …(ii)

From (i) & (ii)

$\displaystyle \frac{1}{2}a t^2 = \frac{1}{2}a’ (2t)^2$

a = 4a’

g sinθ  = 4(g sinθ – μ g cosθ )

4 μ g cosθ = 3 g sinθ

$\displaystyle \mu = \frac{3}{4}tan\theta$