Question:
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:
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:
Show Hint
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
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The Correct Option is D
Solution and Explanation
Concept:
For motion on an inclined plane:
Smooth plane acceleration:
\[
a_1=g\sin\theta
\]
Rough plane acceleration:
\[
a_2=g(\sin\theta-\mu\cos\theta)
\]
Using:
\[
s=\frac12 at^2
\]
For same distance:
\[
a_1T^2=a_2(3T)^2
\]
Step 1: Determine trigonometric ratios. Given: \[ \tan\theta=\frac{9}{16} \] Hence, \[ \sin\theta=\frac9{\sqrt{337}} \] \[ \cos\theta=\frac{16}{\sqrt{337}} \]
Step 2: Use time relation. Since: \[ a_1T^2=a_2(9T^2) \] \[ a_1=9a_2 \] Substitute accelerations: \[ g\sin\theta = 9g(\sin\theta-\mu\cos\theta) \] Cancel \(g\): \[ \sin\theta = 9\sin\theta-9\mu\cos\theta \] \[ 9\mu\cos\theta = 8\sin\theta \] \[ \mu = \frac{8\sin\theta}{9\cos\theta} \] \[ = \frac89\tan\theta \] \[ = \frac89\times\frac9{16} \] \[ =\frac12 \] Therefore, \[ \boxed{0.5} \]
Step 1: Determine trigonometric ratios. Given: \[ \tan\theta=\frac{9}{16} \] Hence, \[ \sin\theta=\frac9{\sqrt{337}} \] \[ \cos\theta=\frac{16}{\sqrt{337}} \]
Step 2: Use time relation. Since: \[ a_1T^2=a_2(9T^2) \] \[ a_1=9a_2 \] Substitute accelerations: \[ g\sin\theta = 9g(\sin\theta-\mu\cos\theta) \] Cancel \(g\): \[ \sin\theta = 9\sin\theta-9\mu\cos\theta \] \[ 9\mu\cos\theta = 8\sin\theta \] \[ \mu = \frac{8\sin\theta}{9\cos\theta} \] \[ = \frac89\tan\theta \] \[ = \frac89\times\frac9{16} \] \[ =\frac12 \] Therefore, \[ \boxed{0.5} \]
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