Q.43 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 43 Solution

Topic: Motion in a Plane · Official NTA Booklet Code M1

Question

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 'θ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by:

✓

Verified Against NTA Final Answer Key

Option (4) : θ = sin^-1[(2g T^2 / (π^2 R))^(1/2)]

📚 Motion in a Plane Medium

📋 Given Parameters & Problem Setup

Circular Radius R

Time Period of Revolution T

Projectile Maximum Height H = 4 R

⚡

The 30-Second Exam Hall Shortcut: ⚡ 2-Step Speed to Angle: v = 2πR/T and H = v² sin²θ / 2g = 4R

⏱️ Save ~25 seconds

$v = \frac{2\pi R}{T} \implies v^2 = \frac{4\pi^2 R^2}{T^2}$ . Substitute into projectile max height: $H = \frac{v^2 \sin^2\theta}{2g} = \frac{4\pi^2 R^2 \sin^2\theta}{2g T^2} = 4R$ . Cancel $4R$ : $\frac{\pi^2 R \sin^2\theta}{2g T^2} = 1 \implies \sin^2\theta = \frac{2g T^2}{\pi^2 R} \implies \theta = \sin^{-1}\left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}$ ! Match Option (4)!

1 Circular motion speed:

v = \frac{2\pi R}{T}

2 Projectile maximum height:

H = \frac{v^2 \sin^2\theta}{2g}

3 Equate

H = 4R

\frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2\theta}{2g} = 4R

\frac{4\pi^2 R^2 \sin^2\theta}{2g T^2} = 4R \implies \frac{\pi^2 R \sin^2\theta}{2g T^2} = 1

5 Solve for

\sin\theta

\sin^2\theta = \frac{2g T^2}{\pi^2 R} \implies \theta = \sin^{-1}\left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}

6 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical uniform circular kinematics coupled with parabolic projectile trajectory equations.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining the Speed from Uniform Circular Motion

A particle completing one full revolution around a circle of radius $R$ in period $T$ travels a distance $2\pi R$ with constant speed:

v = \frac{2\pi R}{T}

Step 2

Formulating the Maximum Height of Projectile Motion

When projected at angle $\theta$ to the horizontal with this initial speed $v$ , the maximum vertical height $H$ reached is:

H = \frac{v^2 \sin^2\theta}{2g}

Step 3

Equating with the Given Maximum Height (H = 4R)

Substitute $v = \frac{2\pi R}{T}$ and $H = 4R$ into the height equation:

\frac{\left(\frac{4\pi^2 R^2}{T^2}\right) \sin^2\theta}{2g} = 4R \implies \frac{4\pi^2 R^2 \sin^2\theta}{2g T^2} = 4R

Step 4

Solving for the Angle of Projection θ

Cancel $4R$ from both sides and isolate $\theta$ :

\frac{\pi^2 R \sin^2\theta}{2g T^2} = 1 \implies \sin^2\theta = \frac{2g T^2}{\pi^2 R} \implies \theta = \sin^{-1}\left[\left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}\right]

⭐

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:

Uses cosine instead of sine and misses the factor of 2 in numerator.

⚠️

Student Fallacy (Why candidates pick this trap): Using horizontal range formula or cosine component.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Inverted ratio and cosine function.

⚠️

Student Fallacy (Why candidates pick this trap): Inversion error.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Inverts the fraction inside the radical and omits the factor of 2.

⚠️

Student Fallacy (Why candidates pick this trap): Inverting numerator and denominator.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact mathematical angle derived from equating projectile height to $4R$ .

💡

Why this is airtight: None; rigorous kinematic derivation.

⚠️

NTA Trap Alert: Fraction Inversion & Trig Function Trap

Candidates frequently arrive at $\sin^2\theta$ and accidentally invert the fraction to $\frac{\pi^2 R}{g T^2}$ (Option 3), or confuse maximum height ( $H \propto \sin^2\theta$ ) with range ( $R_h \propto \sin 2\theta$ ).

🛡️ How to avoid the -1 mark: Cross-multiply carefully:

\sin^2\theta = \frac{2g T^2}{\pi^2 R}

. Take the square root and inverse sine.

📖

Core NEET Syllabus Standard

NCERT Physics · Motion in a Plane Standard Benchmark

Core Scientific Principle Tested:

"Maximum height of a projectile is $H = \frac{v^2 \sin^2\theta}{2g}$ , where speed from circular motion is $v = \frac{2\pi R}{T}$ ."

Syllabus Focus: Projectile Motion and Uniform Circular Motion

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 43 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ 2-Step Speed to Angle: v = 2πR/T and H = v² sin²θ / 2g = 4R

Step-by-Step Rigorous Workflow & Mathematical Proof

Determining the Speed from Uniform Circular Motion

Formulating the Maximum Height of Projectile Motion

Equating with the Given Maximum Height (H = 4R)

Solving for the Angle of Projection θ

360° Option Elimination Engine Key Feature

NTA Trap Alert: Fraction Inversion & Trig Function Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Motion in a Plane

Total Linear Momentum Variation for Two Counter-Rotating Particles on a Circular Wire

Vectors in Uniform Circular Motion

Maximum Height of a Projectile

Projectile Motion from an Accelerating Frame: Velocity and Acceleration upon Release

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