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!

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

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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°.

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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!

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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?

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