Dimensional Analysis JEE 2019: Options A, B and D Explained

What are the correct options in the dimensional analysis JEE 2019 question?

A, B and D are correct in the dimensional analysis JEE 2019 question: energy, force and linear momentum have the proposed dimensions, but power does not. This is Physics, JEE Advanced 2019, Paper 1, Units and Measurements. It is a multiple-correct question, not a JEE Main single-correct MCQ.

The task is to check four proposed dimensions in a system where mass and angular momentum are dimensionless, while length retains its dimension: $[length] = L$

The supplied choices are:

Power actually has dimension: $[P] = L^{- 4}$

The question-bank metadata labels this medium difficulty, with a suggested solving time of 120 seconds. These are practice labels, not an exam-wide timing rule.

How do you derive the new dimension of time?

Time acquires the dimension of length squared because angular momentum must be dimensionless after applying the mass constraint. Keep the ordinary dimensional formulae as your starting point, then use the stipulated conditions to express every dimension through length alone.

First, mass is dimensionless: $[M] = 1$

Here, 1 means dimensionless. It does not mean that the mass is numerically equal to one kilogram.

The standard dimensional formula for angular momentum is:

[J] = M L^{2} T^{- 1}

Angular momentum is also dimensionless, so: $M L^{2} T^{- 1} = 1$

Substitute the mass constraint:

1 \times L^{2} T^{- 1} = 1

$L^{2} T^{- 1} = 1$ Multiply both sides by the time dimension: $L^{2} = T$

Thus:

T = L^{2}

This is a dimensional relationship within the stipulated system, not a numerical claim that one second equals one square metre. You are not changing the definitions of energy or force. You are expressing their usual dimensions under two new constraints.

How do you check energy and force?

Both A and B are correct. Substitute dimensionless mass and the new time dimension into the standard formula for each quantity, then simplify the powers of length.

For energy:

[E] = M L^{2} T^{- 2}

[E] = 1 \times L^{2} \times (L^{2})^{- 2} = L^{2} \times L^{- 4} = L^{- 2}

Raising a power to another power means multiplying the exponents:

(L^{2})^{- 2} = L^{2 \times (- 2)} = L^{- 4}

When multiplying powers with the same base, add their exponents: $2 + 2 (- 2) = 2 - 4 = - 2$

Therefore, option A is correct.

For force: $[F] = M L T^{- 2}$

[F] = 1 \times L \times (L^{2})^{- 2} = L \times L^{- 4} = L^{- 3}

The final exponent calculation is: $1 + (- 4) = - 3$

Therefore, option B is correct. Energy and force contain the same time factor, but energy starts with one extra power of length. Their final dimensions retain that difference.

How do you check power and linear momentum?

Option C is incorrect, while option D is correct. Finding one correct statement does not finish a multiple-correct question: each of the four supplied statements needs its own check.

For power:

[P] = M L^{2} T^{- 3}

Substitute and simplify:

[P] = 1 \times L^{2} \times (L^{2})^{- 3} = L^{2} \times L^{- 6} = L^{- 4}

The time factor contributes an exponent of negative six. Adding the two powers from the length factor gives negative four.

Option C instead states: $L^{- 5}$

These do not agree, so C is incorrect.

For linear momentum: $[p] = M L T^{- 1}$

[p] = 1 \times L \times (L^{2})^{- 1} = L \times L^{- 2} = L^{- 1}

Therefore, D is correct. Here is the option-by-option verdict:

A, energy: agrees. $[E] = L^{- 2}$
B, force: agrees. $[F] = L^{- 3}$
C, power: does not agree.

[P] = L^{- 4} \neq L^{- 5}

D, linear momentum: agrees. $[p] = L^{- 1}$

How can a wrong method produce option C exactly?

A possible method error is to define power as force per unit time. That wrong definition produces exactly the exponent printed in option C. The arithmetic is consistent, but the physical quantity being calculated is not power.

The incorrect starting relation is:

P = \frac{F}{t} (incorrect definition)

It gives:

\frac{[F]}{T} = \frac{L^{- 3}}{L^{2}} = L^{- 5}

That is the dimension of force per unit time, not power. Power is energy transferred per unit time, so for constant power:

P = \frac{Δ E}{Δ t}

Therefore:

[P] = \frac{[E]}{T} = \frac{L^{- 2}}{L^{2}} = L^{- 4}

Write the defining relation before substituting dimensions, especially when comparing force, energy and power. Prefer this check to recognising an option by its exponent: it tests the physics before the algebra.

How do you solve three related Units and Measurements questions?

Use the same substitutions for speed, acceleration and torque. These are original related practice questions, not additional verified JEE PYQs. In every exercise, mass and angular momentum remain dimensionless:

[M] = [J] = 1, T = L^{2}

What is the dimension of speed?

Speed has the dimension of inverse length in this system. Start with distance divided by time, then replace the time dimension:

[v] = L T^{- 1} = L (L^{2})^{- 1} = L \times L^{- 2} = L^{- 1}

Speed and linear momentum share dimensions here because multiplying speed by dimensionless mass introduces no new dimension. Sharing dimensions does not make them the same physical quantity.

What is the dimension of acceleration?

Acceleration has the dimension of inverse length cubed. Its standard formula contains two inverse powers of time:

[a] = L T^{- 2} = L (L^{2})^{- 2} = L \times L^{- 4} = L^{- 3}

Force is mass times acceleration. Since mass contributes a dimensionless factor in this system, force and acceleration share dimensions.

Is torque dimensionless because angular momentum is dimensionless?

No. Torque has the same dimensions as energy, not angular momentum. Direct substitution gives:

[τ] = M L^{2} T^{- 2} = 1 \times L^{2} (L^{2})^{- 2} = L^{2} L^{- 4} = L^{- 2}

Cross-check this using torque as the rate of change of angular momentum:

[τ] = \frac{[J]}{T} = \frac{1}{L^{2}} = L^{- 2}

A dimensionless quantity can have a rate of change that is not dimensionless, because division by time still contributes dimensions. Cover the solutions and redo these three questions, writing the standard formula, substitution and exponent simplification on separate lines.

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For a worked example of the same idea, see How to Study Gravitation JEE: Concepts and Practice Order.

Frequently asked questions

What are the correct options in the dimensional analysis JEE 2019 question?

A, B and D are correct in this JEE Advanced 2019 Paper 1 multiple-correct question. With mass and angular momentum dimensionless, energy has dimension L^-2, force L^-3 and linear momentum L^-1. Option C is incorrect because power has dimension L^-4, not L^-5.

How do you get T = L^2 when mass and angular momentum are dimensionless?

Angular momentum has the standard dimensional formula ML^2T^-1. Setting mass and angular momentum dimensionless gives L^2T^-1 = 1, hence T = L^2. This is a dimensional relationship in the stipulated system, not a claim that one second equals one square metre.

Why is the power option wrong in the JEE 2019 dimensional analysis question?

Power is energy transferred per unit time, so its dimension is L^-2 divided by L^2, which gives L^-4. Option C proposes L^-5. That exponent comes from dividing force by time, which is not the definition of power.

Is torque dimensionless if angular momentum is dimensionless?

No: torque is the rate of change of angular momentum, so its dimension is [J]/T. In this system, [J] = 1 and T = L^2, giving torque the dimension L^-2. A dimensionless quantity need not have a dimensionless rate of change.

Dimensional Analysis JEE 2019: Why A, B and D Are Correct

What are the correct options in the dimensional analysis JEE 2019 question?

How do you derive the new dimension of time?

How do you check energy and force?

How do you check power and linear momentum?

How can a wrong method produce option C exactly?

How do you solve three related Units and Measurements questions?

What is the dimension of speed?

What is the dimension of acceleration?

Is torque dimensionless because angular momentum is dimensionless?

Frequently asked questions

What are the correct options in the dimensional analysis JEE 2019 question?

How do you get T = L^2 when mass and angular momentum are dimensionless?

Why is the power option wrong in the JEE 2019 dimensional analysis question?

Is torque dimensionless if angular momentum is dimensionless?

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