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Q.35 Physics ⭐ 360° Option Elimination Active
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NEET 2021 Physics Question 35 Solution

Topic: Units and Measurements · Official NTA Booklet Code M1
NEET 2021 Physics Question 35: Dimensions of Energy in Terms of Force, Acceleration, and Time

Question

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

✓
Verified Against NTA Final Answer Key
Option (2) : [F] [A] [T^2]
📚 Units and Measurements Easy
📋 Given Parameters & Problem Setup
Fundamental Base Quantities Force [F], Acceleration [A], Time [T]
Target Derived Quantity Energy [E]
⚡

The 30-Second Exam Hall Shortcut: ⚡ Work = Force × Distance, and Distance = 1/2 A T²

⏱️ Save ~20 seconds

Energy has dimensions of Work: [E]=[F]×[Displacement][E] = [F] \times [\text{Displacement}]. From kinematics, displacement s=12at2  ⟹  [s]=[A][T2]s = \frac{1}{2} a t^2 \implies [s] = [A][T^2]. Therefore, [E]=[F][A][T2][E] = [F][A][T^2]! Instant 5-second calculation matching Option (2)!

1 Recall physical definition of work/energy: Energy=Force×Displacement\text{Energy} = \text{Force} \times \text{Displacement}.
2 Express displacement [s][s] in terms of acceleration [A][A] and time [T][T]: s=12at2  ⟹  [s]=[A][T2]s = \frac{1}{2} a t^2 \implies [s] = [A] [T^2].
3 Substitute: [E]=[F][s]=[F][A][T2][E] = [F] [s] = [F] [A] [T^2].
4 Match Option (2).
🛡️
Why You Can 100% Trust This Shortcut: Canonical dimensional analysis and fundamental kinematic definitions.
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulating the Dimensional Equation

Assume the dimensional formula of energy in terms of [F][F], [A][A], and [T][T] is given by:

[E]=[F]x[A]y[T]z[E] = [F]^x [A]^y [T]^z
Step 2

Substituting Fundamental Dimensions in SI Base Units

The standard SI dimensional formulae for the quantities are: - [E]=[M1L2T−2][E] = [M^1 L^2 T^{-2}] - [F]=[M1L1T−2][F] = [M^1 L^1 T^{-2}] - [A]=[M0L1T−2][A] = [M^0 L^1 T^{-2}] - [T]=[T1][T] = [T^1]

[M1L2T−2]=[M1L1T−2]x[L1T−2]y[T1]z=[MxLx+yT−2x−2y+z][M^1 L^2 T^{-2}] = [M^1 L^1 T^{-2}]^x [L^1 T^{-2}]^y [T^1]^z = [M^x L^{x+y} T^{-2x - 2y + z}]
Step 3

Solving for Exponents x, y, and z

Equating powers of MM, LL, and TT on both sides: 1. Power of MM: x=1x = 1 2. Power of LL: x+y=2  ⟹  1+y=2  ⟹  y=1x + y = 2 \implies 1 + y = 2 \implies y = 1 3. Power of TT: −2x−2y+z=−2  ⟹  −2(1)−2(1)+z=−2  ⟹  −4+z=−2  ⟹  z=2-2x - 2y + z = -2 \implies -2(1) - 2(1) + z = -2 \implies -4 + z = -2 \implies z = 2

x=1,y=1,z=2  ⟹  [E]=[F1A1T2]x = 1, \quad y = 1, \quad z = 2 \implies [E] = [F^1 A^1 T^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:

Dimensions of impulse or momentum, not energy ([F][A][T]=[MLT−2][LT−2][T]=[ML2T−3][F][A][T] = [M L T^{-2}][L T^{-2}][T] = [M L^2 T^{-3}]).

⚠️
Student Fallacy (Why candidates pick this trap): Linear time dependence error.
2 2
✓ 100% Correct
Verdict & Scientific Reason:

Exact dimensional formula of energy: [F][A][T2]=[MLT−2][LT−2][T2]=[ML2T−2][F][A][T^2] = [M L T^{-2}][L T^{-2}][T^2] = [M L^2 T^{-2}].

💡
Why this is airtight: None; flawless dimensional equivalence.
3 3
✗ Incorrect
Verdict & Scientific Reason:

Incorrect negative power of time.

⚠️
Student Fallacy (Why candidates pick this trap): Conflating frequency with time.
4 4
✗ Incorrect
Verdict & Scientific Reason:

Dimensions of mass ×\times time ([F/A][T]=[M][T][F/A][T] = [M][T]).

⚠️
Student Fallacy (Why candidates pick this trap): Incorrect exponent inversion.
⚠️

NTA Trap Alert: Power of Time Error Trap

Students who remember [s]=[A][T2][s] = [A][T^2] sometimes write [E]=[F][A][T][E] = [F][A][T] (Option 1) by confusing velocity (v=atv = at) with displacement (s=12at2s = \frac{1}{2}at^2).

🛡️ How to avoid the -1 mark: Check dimensions: [F][A][T2]=(Force)×(Displacement)=Energy[F][A][T^2] = (\text{Force}) \times (\text{Displacement}) = \text{Energy}.
📖

Core NEET Syllabus Standard

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

"The principle of homogeneity of dimensions ensures that physical equations can be converted to any system of fundamental units."

Syllabus Focus: Dimensional Analysis and Its Applications
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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