Laws Of Motion Ques 54
54. 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. The coefficient of friction between the body and the inclined plane is
[1988]
(a) 0.80
(b) 0.75
(c) 0.25
(d) 0.33
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Answer:
Correct Answer: 54.(b)
Solution:
- (b) In presence of friction $a=(g \sin \theta-\mu g \cos \theta)$
$\therefore$ Time taken to slide down the plane
$t_1=\sqrt{\frac{2 s}{a}}=\sqrt{\frac{2 s}{g(\sin \theta-\mu \cos \theta)}}$
In absence of friction, $t_2=\sqrt{\frac{2 s}{g \sin \theta}}$
According to the condition,
$t_1=2 t_2 \quad \therefore t_1^{2}=4 t_2^{2}$
or $\frac{2 s}{g(\sin \theta-\mu \cos \theta)}=\frac{2 s \times 4}{g \sin \theta}$
$\sin \theta=4 \sin \theta-4 \mu \cos \theta$
$\mu=\frac{3}{4} \tan \theta=\frac{3}{4}=0.75$
BETA
