NEET 2025 Physics Friction MCQ Question

To solve the problem, we need to analyze the motion of a body sliding down two inclined surfaces: one smooth and one rough, both inclined at an angle of $45^\circ$ with the horizontal. The key points of this problem involve calculating the time taken for the body to slide down both surfaces and using the relationship between these times to find the coefficient of kinetic friction ($\mu_k$) on the rough surface.

Motion on the Smooth Surface

1. For the smooth incline:
   • The only force acting along the incline is the component of gravitational force, given by: $F = mg \sin(45^\circ)$
   • The acceleration ($a_s$) of the body down the incline is: $a_s = g \sin(45^\circ) = g \cdot \frac{1}{\sqrt{2}}$
   • The time taken ($t_1$) to slide down the length $L$ can be derived from the equation of motion: $L = \frac{1}{2} a_s t_1^2$ Substituting for $a_s$: $L = \frac{1}{2} \left(g \cdot \frac{1}{\sqrt{2}}\right)t_1^2$ Rearranging gives: $t_1^2 = \frac{2L \sqrt{2}}{g}$ Hence, $t_1 = \sqrt{\frac{2L \sqrt{2}}{g}}$

Motion on the Rough Surface

1. For the rough incline:
   • Here, kinetic friction opposes the motion. The frictional force is: $F_{\text{friction}} = \mu_k mg \cos(45^\circ) = \mu_k mg \cdot \frac{1}{\sqrt{2}}$
   • The net force acting down the incline is: $F_{\text{net}} = mg \sin(45^\circ) - F_{\text{friction}} = mg \cdot \frac{1}{\sqrt{2}} - \mu_k mg \cdot \frac{1}{\sqrt{2}}$
   • Thus, the acceleration ($a_r$) is: $a_r = g \sin(45^\circ) - \mu_k g \cos(45^\circ) = g \left(\frac{1}{\sqrt{2}} - \mu_k \cdot \frac{1}{\sqrt{2}}\right)$ Simplifying gives: $a_r = g \cdot \frac{1 - \mu_k}{\sqrt{2}}$
   • The time taken ($t_2$) to slide down the rough incline is given by: $L = \frac{1}{2} a_r t_2^2$ Substituting for $a_r$: $L = \frac{1}{2} \left(g \cdot \frac{1 - \mu_k}{\sqrt{2}}\right)t_2^2$ Rearranging gives: $t_2^2 = \frac{2L\sqrt{2}}{g(1 - \mu_k)}$ Hence, $t_2 = \sqrt{\frac{2L\sqrt{2}}{g(1 - \mu_k)}}$

Relationship Between Times

1. Given Condition:
   • We know from the problem statement that $t_2 = 2t_1$. Squaring both sides gives: $t_2^2 = 4t_1^2$
   • Substituting the expressions for $t_1^2$ and $t_2^2$: $\frac{2L\sqrt{2}}{g(1 - \mu_k)} = 4 \cdot \frac{2L\sqrt{2}}{g}$
   • Simplifying, we can cancel out $L$ and $\sqrt{2}$: $\frac{1}{1 - \mu_k} = 4$
   • Rearranging gives: $1 - \mu_k = \frac{1}{4}$ $\mu_k = 1 - \frac{1}{4} = \frac{3}{4}$

Thus, the coefficient of kinetic friction ($\mu_k$) is:

$\mu_k = 0.75$

Conclusion

The correct answer is D) 0.75.