Q.24 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 24 Solution

Topic: Work, Energy and Power · Official NTA Booklet Code M1

Question

A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively:

✓

Verified Against NTA Final Answer Key

Option (4) : S / 4, \sqrt{3gS / 2}

📚 Work, Energy and Power Easy

📋 Given Parameters & Problem Setup

Initial Release Height S from Earth surface

Initial Velocity 0 (released from rest)

Condition at Intermediate Height h Kinetic energy is three times potential energy: K = 3U

⚡

The 30-Second Exam Hall Shortcut: ⚡ 4-Part Energy Split: U = E/4 $\implies$ h = S/4, v = √(3gS/2)

⏱️ Save ~20 seconds

Total energy $E = mgS = K + U$ . Since $K = 3U$ , we have $4U = mgS \implies U = \frac{1}{4}mgS \implies h = \frac{S}{4}$ ! Kinetic energy is $K = \frac{3}{4}mgS = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{3gS}{2}}$ ! Match Option (4)!

1 Conserve mechanical energy:

E = U_i + K_i = mgS + 0 = mgS

2 At height

h

E = U + K = U + 3U = 4U

3 Equate:

4 mgh = mgS \implies h = \frac{S}{4}

4 Find velocity from kinetic energy:

K = \frac{1}{2}mv^2 = 3U = 3 mg \left(\frac{S}{4}\right) = \frac{3}{4}mgS

v^2 = \frac{3gS}{2} \implies v = \sqrt{\frac{3gS}{2}}

6 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical conservation of mechanical energy in uniform gravitational field.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulating Total Mechanical Energy

At height $S$ above the ground, the particle is released from rest ( $u = 0$ ). Taking the ground as the reference plane for zero gravitational potential energy:

E_{\text{total}} = U_i + K_i = mgS + 0 = mgS

Step 2

Determining the Height from the Earth's Surface

At an unknown height $h$ , potential energy is $U = mgh$ and kinetic energy is $K$ . We are given $K = 3U$ :

E_{\text{total}} = K + U = 3U + U = 4U \implies 4(mgh) = mgS \implies h = \frac{S}{4}

Step 3

Determining the Speed of the Particle

Substitute $h = S/4$ into the expression for kinetic energy $K = \frac{1}{2}mv^2$ :

\frac{1}{2}mv^2 = 3 mgh = 3 mg \left(\frac{S}{4}\right) = \frac{3}{4}mgS \implies v^2 = \frac{3gS}{2} \implies v = \sqrt{\frac{3gS}{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 1

✗ Incorrect

Verdict & Scientific Reason:

Speed lacks the square root (gives $v^2$ instead of $v$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting to take the square root of

v^2

2 2

✗ Incorrect

Verdict & Scientific Reason:

The radical only covers the numerator: $\frac{\sqrt{3gS}}{2} = \sqrt{\frac{3gS}{4}} \neq \sqrt{\frac{3gS}{2}}$ .

⚠️

Student Fallacy (Why candidates pick this trap): Placing the denominator 2 outside the square root.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Incorrect height $S/2$ (which would correspond to $K = U$ , not $K = 3U$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming the condition occurs at the midpoint of fall.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact height and speed derived from energy conservation: $h = S/4$ and $v = \sqrt{\frac{3gS}{2}}$ .

💡

Why this is airtight: None; rigorous algebraic energy solution.

⚠️

NTA Trap Alert: Radical Bracket Scope Trap (Option 2 vs. Option 4)

Notice the typographic trap between Option (2) $\frac{\sqrt{3gS}}{2}$ and Option (4) $\sqrt{\frac{3gS}{2}}$ . In Option 2, the 2 is in the denominator OUTSIDE the radical (giving a factor of $1/4$ under the radical). In Option 4, the 2 is INSIDE the radical.

🛡️ How to avoid the -1 mark: Check radical boundaries carefully:

v = \sqrt{\frac{3gS}{2}}

, so the entire fraction

3gS/2

is under the square root.

📖

Core NEET Syllabus Standard

NCERT Physics · Work, Energy and Power Standard Benchmark

Core Scientific Principle Tested:

"In a conservative gravitational field, total mechanical energy $E = K + U$ is invariant at all points along the trajectory."

Syllabus Focus: Conservation of Mechanical Energy

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 24 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ 4-Part Energy Split: U = E/4 ⟹ \implies⟹ h = S/4, v = √(3gS/2)

Step-by-Step Rigorous Workflow & Mathematical Proof

Formulating Total Mechanical Energy

Determining the Height from the Earth's Surface

Determining the Speed of the Particle

360° Option Elimination Engine Key Feature

NTA Trap Alert: Radical Bracket Scope Trap (Option 2 vs. Option 4)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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The 30-Second Exam Hall Shortcut: ⚡ 4-Part Energy Split: U = E/4 $\implies$ h = S/4, v = √(3gS/2)