Q.148 Physics ⭐ 360° Option Elimination Active

NEET 2020 Physics Question 148 Solution

Topic: System of Particles and Rotational Motion · Official NTA Booklet Code E1

Question

Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of:

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Verified Against NTA Final Answer Key

Option (3) : 67 cm

📚 System of Particles and Rotational Motion Easy

📋 Given Parameters & Problem Setup

Mass 1 (m1) 5 kg (located at origin x = 0)

Mass 2 (m2) 10 kg (located at x = L = 1 m = 100 cm)

Rod Length (L) 1 m = 100 cm (mass of rod is negligible)

Question Objective Find the distance of centre of mass from 5 kg particle

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The 30-Second Exam Hall Shortcut: ⚡ "Inverse Mass Ratio: x1 = [m2 / (m1 + m2)] × L = (10 / 15) × 100 cm = 66.7 cm ≈ 67 cm!"

⏱️ 15 seconds

The centre of mass divides the distance between two masses in the inverse ratio of their masses: $\frac{x_1}{x_2} = \frac{m_2}{m_1} = \frac{10}{5} = 2$ . Thus, the 5 kg particle is twice as far from the centre of mass as the 10 kg particle. Distance from 5 kg particle is $x_{cm} = \frac{m_2}{m_1 + m_2} L = \frac{10}{15} \times 100\text{ cm} = \frac{2}{3} \times 100 = 66.67\text{ cm} \approx 67\text{ cm}$ . Mark Option (3) in 5 seconds!

1 Formula: x_cm = (m1 x1 + m2 x2) / (m1 + m2).

2 Choose 5 kg at origin (x1 = 0) and 10 kg at x2 = 100 cm.

3 Substitute: x_cm = (5 × 0 + 10 × 100) / (5 + 10) = 1000 / 15.

4 Divide: 1000 / 15 = 200 / 3 = 66.67 cm ≈ 67 cm.

5 Select Option (3) 67 cm.

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Why You Can 100% Trust This Shortcut: NCERT Physics · System of Particles and Rotational Motion (Centre of Mass of a Two-Particle System)

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Define Coordinate System

Place the origin ( $x = 0$ ) at the position of the 5 kg particle ( $m_1 = 5\text{ kg}$ ). The rigid light rod extends along the x-axis, placing the 10 kg particle ( $m_2 = 10\text{ kg}$ ) at coordinate $x = L = 1\text{ m} = 100\text{ cm}$ .

Step 2

Apply Centre of Mass Formula

For a two-particle discrete mass system along a straight line: $x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Step 3

Substitute Given Values

$x_{\text{cm}} = \frac{(5\text{ kg} \times 0) + (10\text{ kg} \times 100\text{ cm})}{5\text{ kg} + 10\text{ kg}}$ $x_{\text{cm}} = \frac{1000\text{ kg}\cdot\text{cm}}{15\text{ kg}} = \frac{200}{3}\text{ cm} = 66.666...\text{ cm}$

Step 4

Round to Nearest Integer

Rounding to the nearest whole integer gives $67\text{ cm}$ , matching Option (3).

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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; 33 cm ( $\frac{1}{3}L$ ) is the distance of the centre of mass from the HEAVIER 10 kg particle ( $100 - 67 = 33\text{ cm}$ ), not from the 5 kg particle.

⚠️

Student Fallacy (Why candidates pick this trap): Measuring distance from the wrong end of the rod.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; 50 cm would be the midpoint if both particles had equal masses.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming geometric midpoint instead of mass-weighted centre.

3 C

✓ 100% Correct

Verdict & Scientific Reason:

Factually and mathematically exact; $x_{\text{cm}} = \frac{10}{15} \times 100\text{ cm} = 66.67\text{ cm} \approx 67\text{ cm}$ .

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Why this is airtight: None; clean centre of mass calculation.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; corresponds to an erroneous mass ratio of 4 : 1.

⚠️

Student Fallacy (Why candidates pick this trap): Arithmetic mistake.

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NTA Trap Alert: From 5 kg vs From 10 kg Trap

Always check which mass the distance is measured from! - Distance from 5 kg (lighter): $67\text{ cm}$ (closer to 10 kg). - Distance from 10 kg (heavier): $33\text{ cm}$ . Option (1) has 33 cm as a classic distractor!

🛡️ How to avoid the -1 mark: Centre of mass is always closer to the heavier mass! Distance from 5 kg must be

> 50\text{ cm}

-> 67 cm!

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Core NEET Syllabus Standard

Physics · System of Particles and Rotational Motion Standard Benchmark

Core Scientific Principle Tested:

Syllabus Focus: Centre of Mass of Two-Particle System: x_cm = m2 L / (m1 + m2)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2020 Physics Question 148 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ "Inverse Mass Ratio: x1 = [m2 / (m1 + m2)] × L = (10 / 15) × 100 cm = 66.7 cm ≈ 67 cm!"

Step-by-Step Rigorous Workflow & Mathematical Proof

Define Coordinate System

Apply Centre of Mass Formula

Substitute Given Values

Round to Nearest Integer

360° Option Elimination Engine Key Feature

NTA Trap Alert: From 5 kg vs From 10 kg Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — System of Particles and Rotational Motion

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