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 $