Question

Starting from rest, a body slides down a ${45}^{\circ }$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction (with same inclination). The coefficient of friction between the body and the inclined plane is

• A. $0.33$
• B. $0.25$
• C. $0.75$
• D. $0.80$

Answer 1

The correct option is C $0.75$

Acceleration without friction $a_w=\text{g sin}\theta$
acceleration with friction $a_f=\text{g sin}\theta -\mu \text{g cos}\theta$
Now time taken $t=\sqrt{\frac{2h}{a}}$
Let $T$ time taken in the absence of friction,
$T_f$ time taken in the presence of friction.

Given that $T_f=2T$

$\Rightarrow \frac{T}{T_f}=\sqrt{\frac{a_f}{a_w}}=\sqrt{\frac{gsin\theta -\text{mg cos}\theta }{\text{g sin}\theta }}$

$\Rightarrow \frac{1}{2}=\sqrt{1-\mu cot\theta }$
Squaring on both sides and putting the value of $\cot \theta =\cot {45}^{\circ }=1$,
we get $\mu =\frac{3}{4}=0.75$