NEET 2025 Q1: Time ratio for sliding down rough vs. smooth inclines — Full solution + Why they asked this

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1. TITLE

NEET 2025 Q1: Time ratio for sliding down rough vs. smooth inclines — Full solution + Why they asked this

2. THE ACTUAL NEET QUESTION

NEET 2025 — Question
There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_{k})$ between the object and the rough surface is close to
A) 0.5
B) 0.75
C) 0.25
D) 0.40

3. OFFICIAL ANSWER

NEET Answer Key: (B)

4. COMPLETE SOLUTION

Step-by-step calculation:
1. Let the time taken to slide down the smooth surface be $t_{smooth}$ and the rough surface be $t_{rough}$. We are given $t_{rough} = 2t_{smooth}$.
2. For a body sliding down a perfectly smooth inclined plane, the downward acceleration is:
$a_{smooth} = g \sin \theta$
3. For a body sliding down a rough inclined plane, the downward acceleration is opposed by kinetic friction ($f_k = \mu_k N = \mu_k mg \cos \theta$):
$a_{rough} = g \sin \theta - \mu_k g \cos \theta$
4. Using the kinematic equation $S = ut + \frac{1}{2}at^2$, since initial velocity $u = 0$ and the distance $S = L$ is the same for both, the time $t$ is inversely proportional to the square root of acceleration:
$t \propto \frac{1}{\sqrt{a}}$
5. Therefore, the time ratio is:
$\frac{t_{rough}}{t_{smooth}} = \sqrt{\frac{a_{smooth}}{a_{rough}}} = 2$
6. Squaring both sides:
$\frac{g \sin \theta}{g \sin \theta - \mu_k g \cos \theta} = 4$
7. Substituting $\theta = 45^{\circ}$ (where $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$):
$\frac{1/\sqrt{2}}{1/\sqrt{2} - \mu_k(1/\sqrt{2})} = 4$
8. Canceling $\frac{1}{\sqrt{2}}$ from the numerator and denominator:
$\frac{1}{1 - \mu_k} = 4 \implies 1 - \mu_k = 0.25$
9. Solving for $\mu_k$:
$\mu_k = 1 - 0.25 = 0.75$

Formula used:
- $a = g(\sin \theta - \mu_k \cos \theta)$
- $S = \frac{1}{2}at^2$

5. WHY NEET ASKED THIS

What concept were they testing?
The examiner was testing the integration of Kinematics (equations of motion) with the Laws of Motion (resolving forces on an inclined plane and calculating kinetic friction).

What's the common mistake they expected students to make?
The most common mistake is forgetting the square root in the time-acceleration relationship. Students often write $\frac{t_{rough}}{t_{smooth}} = \frac{a_{smooth}}{a_{rough}} = 2$, which leads to $\mu_k = 0.5$.

Which option was the "trap"?
Option (A) 0.5 is the trap answer exactly for those who forget to square the time ratio before solving for $\mu_k$!

6. TREND ANALYSIS

PATTERN ALERT:

Prediction for NEET 2026:
Expect a variation where the coefficient of friction ($\mu_k$) is provided, and you must calculate the exact time it takes to slide down, or you will be asked to find the angle of inclination ($\theta$) given the time ratio and $\mu_k$.

7. SIMILAR QUESTIONS TO PRACTICE

If you understood this, you can solve:
- NEET 1988 Q76 (Exact conceptual match)
- NEET 2015 Q21 (Calculating friction from distance and time on an incline)
- NEET 2013 Q28 (Smooth vs. rough sections on the same incline)

8. QUICK REVISION FOR THIS TYPE

For Time ratio for sliding down rough vs. smooth inclines questions, always:

• First check: Is the body starting from rest and covering the exact same distance?
• Then apply: The universal shortcut formula for these specific problems: (Note: This standard shortcut is mathematically derived directly from the steps in)
  $\mu_k = \tan \theta \left[ 1 - \left(\frac{1}{n}\right)^2 \right]$
  (Where '$n$' is the multiple of time taken on the rough incline compared to the smooth one).
  For this 2025 PYQ: $\mu_k = \tan(45^{\circ}) \left[ 1 - (\frac{1}{2})^2 \right] = 1 \times [1 - 0.25] = 0.75$.
• Watch out for: The angle $\theta$. If $\theta$ is 30° or 60°, $\tan \theta$ will fundamentally change the final multiplier. Don't blindly memorize the "0.75" answer!

9. NCERT BASE

This question tests: NCERT Class 11, Chapter 4 (Laws of Motion), pages discussing Kinetic Friction and resolution of forces on inclined planes.

10. CTA

All PYQs from this chapter → Laws of Motion Guide
Practice similar questions → Mock exams