Question:hard

A body is placed at the top of an inclined plane of angle of inclination \( \tan^{-1}\left(\frac{9}{16}\right) \). If the plane is smooth, the body reaches the bottom in time \(T\) and if the plane is rough, it takes time \(3T\) to reach the bottom of the plane, then the coefficient of kinetic friction between the body and the rough inclined plane is:

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For the same displacement under different accelerations: \[ a_1t_1^2=a_2t_2^2 \] This shortcut saves a lot of algebra in kinematics problems.
Updated On: Jun 17, 2026
  • \(0.2\)
  • \(0.3\)
  • \(0.4\)
  • \(0.5\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Set up the two cases.
The same body slides down the same slope twice. First the slope is smooth, then it is rough. The slope length is the same both times, but the rough one is slower because friction fights the motion.

Step 2: Acceleration on each slope.
On the smooth slope only gravity acts along the incline. \[ a_1 = g\sin\theta \] On the rough slope friction is subtracted. \[ a_2 = g(\sin\theta - \mu\cos\theta) \]
Step 3: Use the same distance idea.
Starting from rest, distance is $s = \tfrac12 a t^2$. Since the distance is equal in both cases, and times are $T$ and $3T$, \[ a_1 T^2 = a_2 (3T)^2 \] This gives $a_1 = 9a_2$.
Step 4: Plug in the accelerations.
\[ g\sin\theta = 9\,g(\sin\theta - \mu\cos\theta) \] Cancel $g$ from both sides.
Step 5: Solve for the friction coefficient.
\[ \sin\theta = 9\sin\theta - 9\mu\cos\theta \] \[ 9\mu\cos\theta = 8\sin\theta \] \[ \mu = \frac89 \tan\theta \]
Step 6: Use the given angle.
Here $\tan\theta = \tfrac{9}{16}$. \[ \mu = \frac89 \times \frac{9}{16} = \frac12 = 0.5 \] \[ \boxed{0.5} \]
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