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Q.48 Physics · Section B ⭐ 360° Option Elimination Active
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NEET 2024 Physics Question 48 Solution

Topic: Units and Measurements · Official NTA Booklet Code T1
NEET 2024 Physics Question 48: Identifying Dimensionless Combinations from an Equation of Force

Question

A force defined by F = αt² + βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:

✓
Verified Against NTA Final Answer Key
Option (4) : αt / β
📚 Units and Measurements Easy
📋 Given Parameters & Problem Setup
Force equation F = αt² + βt
Constants α and β are dimensional constants
Time variable t
⚡

The 30-Second Exam Hall Shortcut: ⚡ Dimensional Ratio Inspection

⏱️ Save ~35 seconds

Both terms have dimension of Force: [αt²] = [βt] -> [αt²] / [βt] = [αt / β] = 1 (Dimensionless)!

1 Principle of Dimensional Homogeneity: In F = αt² + βt, every term added has identical dimensions of Force: [αt²] = [βt] = [F].
2 Divide the first term by the second term: (αt²) / (βt) = αt / β.
3 The ratio of two quantities having identical dimensions is strictly DIMENSIONLESS: [αt / β] = [F] / [F] = 1 = [M⁰L⁰T⁰].
4 Option (4) in 3 seconds!
🛡️
Why You Can 100% Trust This Shortcut: Principle of Dimensional Homogeneity (Terms added in a physical equation have identical dimensions).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Principle of Dimensional Homogeneity

According to Fourier’s principle of dimensional homogeneity, physical quantities can be added or equated only if they have the same dimensions. In the force equation F = αt² + βt, both terms on the right must independently have dimensions of force [M L T⁻²]:

[αt2]=[F]=[MLT−2],[βt]=[F]=[MLT−2][\alpha t^2] = [F] = [M L T^{-2}], \quad [\beta t] = [F] = [M L T^{-2}]
Step 2

Determination of Dimensions of α and β

Solving for the dimensional formulas of α and β:

[α]=[F][t2]=[MLT−2][T2]=[MLT−4][\alpha] = \frac{[F]}{[t^2]} = \frac{[M L T^{-2}]}{[T^2]} = [M L T^{-4}]
Step 3

Dimensions of β

Solving for the dimensions of β:

[β]=[F][t]=[MLT−2][T]=[MLT−3][\beta] = \frac{[F]}{[t]} = \frac{[M L T^{-2}]}{[T]} = [M L T^{-3}]
Step 4

Testing the Ratio αt / β

Evaluating the dimensions of the expression αt / β:

[αtβ]=[α][t][β]=[MLT−4][T][MLT−3]=[MLT−3][MLT−3]=[M0L0T0]\left[\frac{\alpha t}{\beta}\right] = \frac{[\alpha] [t]}{[\beta]} = \frac{[M L T^{-4}] [T]}{[M L T^{-3}]} = \frac{[M L T^{-3}]}{[M L T^{-3}]} = [M^0 L^0 T^0]
⭐

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:

[αβt] = [M² L² T⁻⁶], not dimensionless.

⚠️
Student Fallacy (Why candidates pick this trap): Multiplying dimensions instead of taking ratio.
2 (2)
✗ Incorrect
Verdict & Scientific Reason:

[αβ / t] = [M² L² T⁻⁸], not dimensionless.

⚠️
Student Fallacy (Why candidates pick this trap): Dimensional mismatch.
3 (3)
✗ Incorrect
Verdict & Scientific Reason:

[βt / α] = [T²], has dimensions of time squared.

⚠️
Student Fallacy (Why candidates pick this trap): Inverting the ratio.
4 (4)
✓ 100% Correct
Verdict & Scientific Reason:

Rigorous dimensional cancellation yields [M⁰L⁰T⁰].

💡
Why this is airtight: None.
⚠️

NTA Trap Alert: Detailed Algebraic Derivation vs Direct Ratio Inspection

Students waste 2 minutes calculating [α] and [β] separately and then testing all four options one by one.

🛡️ How to avoid the -1 mark: Direct shortcut: Since αt² and βt both equal Force, their ratio (αt²)/(βt) = αt/β MUST be dimensionless!
📖

Core NEET Syllabus Standard

Physics Core General Principles · Units and Measurements Standard Benchmark
Core Scientific Principle Tested:

"Terms added together in an equation of physics must have identical dimensions; their ratio is always dimensionless."

Syllabus Focus: Dimensional Analysis and Its Applications
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
Both parts equal Force in size: Divide them out and dimensions die!

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