Q.22 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 22 Solution

Topic: System of Particles and Rotational Motion · Official NTA Booklet Code 45

Question

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take g = 10 m/s²)

✓

Verified Against NTA Final Answer Key

Option (2) : 100\sqrt{3} N

📚 System of Particles and Rotational Motion Medium

📋 Given Parameters & Problem Setup

Rod Mass (m) 20 kg (Weight W = mg = 200 N)

Rod Length (L) 5 m

Inclination Angle 60° with the smooth vertical wall

Wall & Floor Characteristics Vertical wall is frictionless (smooth); Floor is rough horizontal

⚡

The 30-Second Exam Hall Shortcut: ⚡ Wall Reaction & Friction Torque Shortcut

⏱️ Save ~45 seconds

Angle with wall is 60° ⟹ Angle with floor is 30°! Torque about base: N_w (L cos 60°) = mg (L/2 sin 60°) ⟹ N_w = (mg/2) tan 60° = (200/2) √3 = 100√3 N = Friction!

1 Notice: The problem says "making an angle of 60° with IT" (with the vertical wall).

2 Floor friction f balances the horizontal normal reaction from the wall N_w: f = N_w.

3 Take torque about base contact point on floor:

4 Torque of N_w = N_w × L cos 60° (counter-clockwise).

5 Torque of weight mg = mg × (L/2) sin 60° (clockwise).

6 Equate: N_w = (mg / 2) × (sin 60° / cos 60°) = (mg / 2) tan 60° = 100 × √3 = 100√3 N.

7 Since f = N_w, friction force = 100√3 N.

8 Select Option (2) in under 20 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Rotational Equilibrium of Rigid Bodies (NCERT Physics · System of Particles and Rotational Motion: Στ = 0).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Force Balance on the Leaning Rod

Four forces act on the uniform rod in equilibrium: (1) Normal reaction at the smooth wall N_w acting horizontally to the right. (2) Static friction force f from the rough floor acting horizontally to the left. (3) Normal reaction from the floor N_f acting vertically upward. (4) Weight of the rod W = mg = 20 × 10 = 200 N acting downward at the midpoint.

\sum F_x = 0 \implies f = N_w, \quad \sum F_y = 0 \implies N_f = mg = 200\text{ N}

• The wall is smooth, so there is no vertical friction at the wall.

Step 2

Rotational Equilibrium About Base Contact Point

Let the rod have length L. The angle with the vertical wall is θ = 60°. The top of the rod is at height h = L cos 60°, and the center of mass is at horizontal distance x_cm = (L/2) sin 60° from the base. Taking torque about the base point on the floor:

\sum \tau_{base} = 0 \implies N_w \cdot (L \cos 60^\circ) = mg \cdot \left(\frac{L}{2} \sin 60^\circ\right)

• N_w = (mg / 2) tan 60°.

Step 3

Calculate Wall Reaction and Friction Force

Substitute the given values mg = 200 N and tan 60° = √3:

N_w = \frac{200\text{ N}}{2} \times \sqrt{3} = 100\sqrt{3}\text{ N} \implies f = N_w = 100\sqrt{3}\text{ N}

• Thus the friction force exerted by the floor on the rod is 100√3 N, matching Option (2).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Forgets the tan 60° = √3 factor (takes mg/2 = 100 N).

⚠️

Student Fallacy (Why candidates pick this trap): Omitting the angular geometric lever arm factor.

2 B

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. f = N_w = (mg/2) tan 60° = 100√3 N.

💡

Why this is airtight: None; rigorous rigid body torque balance.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. 200 N is the vertical normal reaction from the floor N_f, not horizontal friction.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing vertical normal reaction with horizontal friction force.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Forgets the factor of 1/2 from the center of mass position (L/2).

⚠️

Student Fallacy (Why candidates pick this trap): Treating weight as acting at the top of the rod instead of center of mass.

⚠️

NTA Trap Alert: Angle Reference Ambiguity (Wall vs. Floor)

The question states: "making an angle of 60° with IT" (the vertical wall). If a student misreads this as 60° with the floor, they calculate (mg/2) tan 30° = 100/√3 N. Always read the reference surface carefully!

🛡️ How to avoid the -1 mark: Identify what "it" refers to: smooth vertical wall ⟹ angle with wall = 60°.

📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · System of Particles and Rotational Motion Standard Benchmark

Core Scientific Principle Tested:

"Translational equilibrium (Σ F = 0) and rotational equilibrium (Σ τ = 0) uniquely determine support reactions."

Syllabus Focus: Equilibrium of a Rigid Body

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Torque about base: N_w = (mg/2) tan(θ_wall) = (200/2) √3 = 100√3 N!

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 180-minute timed NTA simulator.

Launch NEET 2025 CBT Test →

← Previous (Q.21) All Physics Questions Grid Next (Q.23) →

🔗 Related Questions — System of Particles and Rotational Motion

All 13 System of Particles and Rotational Motion PYQs →

NEET 2026 Q27

Moment of Inertia Difference for Attached Solid Spheres via Parallel Axis Theorem

View Solution →

NEET 2026 Q31

Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2025 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2025 Physics Question 22 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Wall Reaction & Friction Torque Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Force Balance on the Leaning Rod

Rotational Equilibrium About Base Contact Point

Calculate Wall Reaction and Friction Force

360° Option Elimination Engine Key Feature

NTA Trap Alert: Angle Reference Ambiguity (Wall vs. Floor)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — System of Particles and Rotational Motion

Moment of Inertia Difference for Attached Solid Spheres via Parallel Axis Theorem

Angular Momentum of a Fixed-Center Rotating Disc About Arbitrary Points

Velocity of Points on a Rolling Wheel

Length of a Thin Rod from its Central Moment of Inertia

Solve 4 NEET Questions a Day — One per Subject