Q.48 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 48 Solution

Topic: System of Particles and Rotational Motion · Official NTA Booklet Code M1

Question

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. (g = 10 m/s²)

✓

Verified Against NTA Final Answer Key

Option (4) : 1/12 kg

📚 System of Particles and Rotational Motion Medium

📋 Given Parameters & Problem Setup

Length of Uniform Rod (L) 200 cm

Mass of Rod (M_rod) 500 g = 0.5 kg

Fulcrum (Wedge) Position 40 cm mark

Known Mass (m_1) 2 kg at 20 cm mark

Unknown Mass (m) Suspended at 160 cm mark

⚡

The 30-Second Exam Hall Shortcut: ⚡ Torque Balance about Wedge at 40 cm

⏱️ Save ~25 seconds

Fulcrum is at $40\text{ cm}$ . Center of mass of the 200 cm rod is at $100\text{ cm}$ (distance $= 60\text{ cm}$ to right, mass $= 0.5\text{ kg}$ ). Counter-clockwise torque: $2\text{ kg} \times (40 - 20) = 40\text{ kg}\cdot\text{cm}$ . Clockwise torque: $(0.5 \times 60) + m \times (160 - 40) = 30 + 120m$ . Equate: $40 = 30 + 120m \implies 120m = 10 \implies m = 1/12\text{ kg}$ ! Match Option (4)!

1 Locate the center of mass of uniform 200 cm rod: at

100\text{ cm}

mark. Weight of rod is

0.5 g

2 Fulcrum is at

40\text{ cm}

mark.

3 Lever arms from fulcrum: - 2 kg mass at 20 cm: arm

= 40 - 20 = 20\text{ cm}

(left, counter-clockwise). - Rod weight (0.5 kg) at 100 cm: arm

= 100 - 40 = 60\text{ cm}

(right, clockwise). - Unknown mass

m

at 160 cm: arm

= 160 - 40 = 120\text{ cm}

(right, clockwise).

4 Balance torques:

2 \times 20 = (0.5 \times 60) + (m \times 120)

40 = 30 + 120m \implies 120m = 10 \implies m = \frac{10}{120} = \frac{1}{12}\text{ kg}

6 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical principle of moments for static equilibrium of a rigid body.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Identifying Forces and Points of Action

For a uniform rod of length $L = 200\text{ cm}$ and mass $M = 500\text{ g} = 0.5\text{ kg}$ : - Center of mass of the rod is at its geometric center: $x_{\text{cm}} = 100\text{ cm}$ . The gravitational force of the rod $M g = 0.5 g$ acts downward at this point. - The wedge (fulcrum) is located at $x_w = 40\text{ cm}$ . - A mass $m_1 = 2\text{ kg}$ is suspended at $x_1 = 20\text{ cm}$ . - An unknown mass $m$ is suspended at $x_2 = 160\text{ cm}$ .

Step 2

Calculating Lever Arms with Respect to the Fulcrum

Measuring perpendicular distances from the fulcrum ( $x_w = 40\text{ cm}$ ): - For $m_1 = 2\text{ kg}$ : $d_1 = 40 - 20 = 20\text{ cm}$ (to the left, producing CCW torque) - For the rod weight: $d_{\text{rod}} = 100 - 40 = 60\text{ cm}$ (to the right, producing CW torque) - For unknown mass $m$ : $d_2 = 160 - 40 = 120\text{ cm}$ (to the right, producing CW torque)

Step 3

Applying the Principle of Moments (Rotational Equilibrium)

For the rod to remain in rotational equilibrium, the sum of counter-clockwise moments must equal the sum of clockwise moments about the wedge:

\Sigma \tau = 0 \implies m_1 g d_1 = M g d_{\text{rod}} + m g d_2

Step 4

Solving for Unknown Mass m

Cancel $g$ and substitute numerical values:

2 \times 20 = (0.5 \times 60) + (m \times 120) \implies 40 = 30 + 120 m \implies 120 m = 10 \implies m = \frac{10}{120} = \frac{1}{12}\text{ kg}

⭐

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 1

✗ Incorrect

Verdict & Scientific Reason:

Result of neglecting the weight of the rod entirely ( $2 \times 20 = m \times 120 \implies m = 40/120 = 1/3$ , or other algebraic error).

⚠️

Student Fallacy (Why candidates pick this trap): Ignoring the mass of the rod.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of treating the rod as massless ( $40 = 120m \implies m = 1/3\text{ kg}$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming a massless meter stick.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Placing center of mass of the rod at 50 cm instead of 100 cm (assuming 100 cm total rod length instead of 200 cm).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming rod length is 100 cm instead of reading 200 cm.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact mass required for equilibrium: $m = \frac{40 - 30}{120} = \frac{1}{12}\text{ kg}$ .

💡

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

⚠️

NTA Trap Alert: Rod Length (200 cm vs. 100 cm) Trap

Students automatically think "meter rule" and place the center of mass at $50\text{ cm}$ . The problem explicitly states the rod length is $200\text{ cm}$ , placing the center of mass at $100\text{ cm}$ mark ( $60\text{ cm}$ from the wedge).

🛡️ How to avoid the -1 mark: Read the prompt carefully: Length is

200\text{ cm} \implies

Center of mass is at

100\text{ cm}

📖

Core NEET Syllabus Standard

NCERT Physics · System of Particles and Rotational Motion Standard Benchmark

Core Scientific Principle Tested:

"A rigid body is in mechanical equilibrium when both net external force and net external torque about any pivot point are zero ( $\Sigma \tau = 0$ )."

Syllabus Focus: Equilibrium of a Rigid Body (Principle of Moments)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 48 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Torque Balance about Wedge at 40 cm

Step-by-Step Rigorous Workflow & Mathematical Proof

Identifying Forces and Points of Action

Calculating Lever Arms with Respect to the Fulcrum

Applying the Principle of Moments (Rotational Equilibrium)

Solving for Unknown Mass m

360° Option Elimination Engine Key Feature

NTA Trap Alert: Rod Length (200 cm vs. 100 cm) Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — System of Particles and Rotational Motion

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