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 $