Dimensional Analysis JEE 2020: Why A and B Are Correct

What is the dimensional analysis JEE 2020 question, and which options are correct?

A and B are correct in this dimensional analysis JEE 2020 question from JEE Advanced 2020, Paper 2, Physics, Units and Measurements. More than one option can be correct.

The question bank classifies it as medium difficulty and gives 90 seconds as its expected solving time. These are the bank’s labels, not an official JEE difficulty rating or an exam-imposed time limit.

The unit system expresses the dimensions of position, speed, acceleration, linear momentum and force as powers of a single quantity:

[position] = [X]^{α}, [speed] = [X]^{β}

[acceleration] = [X]^{p}, [linear momentum] = [X]^{q}, [force] = [X]^{r}

The four candidate relations are:

A $α + p = 2 β$
B $p + q - r = β$
C $p - q + r = α$
D $p + q + r = β$

Why is option A correct?

Acceleration multiplied by position has the dimensions of speed squared, which proves A. Square brackets denote dimensions, not numerical values. The following symbols are exponents, not the physical quantities themselves: $α, β, p, q, r$

Multiplication adds exponents, division subtracts exponents, and raising a quantity to a power multiplies its exponent. Start with the official solution’s dimensional identity:

[acceleration] [position] = [speed]^{2}

Check it using standard dimensions:

(L T^{- 2}) (L) = L^{2} T^{- 2} = (L T^{- 1})^{2}

Substitute the given powers and combine exponents:

[X]^{p} [X]^{α} = {([X]^{β})}^{2}

[X]^{p + α} = [X]^{2 β}

$p + α = 2 β$ A is correct. This is a dimensional identity, not a claim that acceleration times position always equals speed squared numerically. Matching dimensions alone does not establish a physical equation or determine its numerical factors.

Why is option B correct?

Acceleration times linear momentum divided by force has speed dimensions, so B is correct. Translate the exponent combination before cancelling anything: subtracting the force exponent places force in the denominator.

p + q - r ⟷ \frac{[acceleration] [linear momentum]}{[force]}

The required dimensions are:

\begin{aligned} [\text{acceleration}]&=LT^{-2}\\ [\text{linear momentum}]&=MLT^{-1}\\ [\text{force}]&=MLT^{-2} \end{aligned}

Multiply the numerator first, then divide:

\frac{(L T^{- 2}) (M L T^{- 1})}{M L T^{- 2}} = \frac{M L^{2} T^{- 3}}{M L T^{- 2}}

= M^{1 - 1} L^{2 - 1} T^{- 3 - (- 2)} = L T^{- 1}

The result is the dimension of speed. The denominator’s negative time exponent is subtracted, not copied.

Returning to the given notation:

\frac{[X]^{p} [X]^{q}}{[X]^{r}} = [X]^{p + q - r} = [X]^{β}

p + q - r = β

For a numerical illustration, take a body with the following mass, speed and acceleration, using net force:

m = 2 k g, v = 4 m s^{- 1}, a = 3 m s^{- 2}

P = m v = 8 k g m s^{- 1}, F = m a = 6 N

\frac{a P}{F} = \frac{3 \times 8}{6} m s^{- 1} = 4 m s^{- 1}

This illustrates the quotient for one body. The dimensional cancellation, not this numerical example, proves B.

Why are options C and D incorrect?

C leaves an unwanted inverse-time-cubed factor; D fails to match speed in mass, length and time powers. Neither is a generally valid dimensional identity. Finding two correct choices does not settle the remaining options in a multiple-correct question.

For C, subtracting the momentum exponent puts momentum in the denominator:

p - q + r ⟷ \frac{[acceleration] [force]}{[linear momentum]}

Evaluate and compare with position:

\frac{(L T^{- 2}) (M L T^{- 2})}{M L T^{- 1}} = \frac{M L^{2} T^{- 4}}{M L T^{- 1}} = L T^{- 3}

[position] = L, L T^{- 3} \neq L

C is incorrect. Cancelling mass is not enough: the remaining time dimension must also match.

For D, all three exponents are added, so all three dimensions must be multiplied:

p + q + r ⟷ [acceleration] [linear momentum] [force]

(L T^{- 2}) (M L T^{- 1}) (M L T^{- 2}) = M^{2} L^{3} T^{- 5}

Compare with speed:

[speed] = L T^{- 1}, M^{2} L^{3} T^{- 5} \neq L T^{- 1}

D is incorrect. Unlike B, it multiplies by force instead of dividing by it.

Select A and B.

How can losing the denominator turn option B into option D?

Adding the force exponent changes division by force into multiplication by force. A possible faulty step is to identify acceleration times momentum divided by force correctly, recognise speed dimensions, but then write: $p + q + r = β$

Compare the incorrect and correct translations:

\underset{incorrect translation}{\underset{⏟}{\frac{[X]^{p} [X]^{q}}{[X]^{r}} \neq [X]^{p + q + r}}} \underset{correct translation}{\underset{⏟}{\frac{[X]^{p} [X]^{q}}{[X]^{r}} = [X]^{p + q - r}}}

The plus sign tests a different physical product, not the quotient you checked. Its dimensions are those calculated under D.

Keep the fraction bar until the last step. Write the physical product or quotient first, substitute dimensions or powers second, and combine exponents third. Prefer this written sequence to mentally converting a phrase into an exponent sum.

How do you apply this method to three related questions?

Translate time, mass and work into dimensional quotients or products before combining exponents. These are original related practice questions, not additional verified JEE PYQs.

1. If time has the following dimensions, express its exponent using the position and speed exponents. $[time] = [X]^{t}$

Time has the dimensions of position divided by speed:

[time] = \frac{[position]}{[speed]}

[X]^{t} = [X]^{α - β} \Rightarrow t = α - β

2. If mass has the following dimensions, express its exponent using the momentum and speed exponents. $[mass] = [X]^{m}$

Divide linear momentum by speed:

[mass] = \frac{[linear momentum]}{[speed]}

[X]^{m} = \frac{[X]^{q}}{[X]^{β}} \Rightarrow m = q - β

3. If work has the following dimensions, express its exponent in two ways. $[work] = [X]^{w}$

Force times position gives the first answer:

[work] = [force] [position] \Rightarrow w = r + α

Linear momentum times speed gives the second:

[work] = [linear momentum] [speed] \Rightarrow w = q + β

Verify agreement using A and B:

α = 2 β - p, r = p + q - β

r + α = (p + q - β) + (2 β - p) = q + β

Cover these answers and redo all three. Write the physical quotient or product before combining the exponents.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with How to Study Biomolecules JEE: A Six-Step NCERT Plan.

Frequently asked questions

Which options are correct in the JEE Advanced 2020 dimensional analysis question?

A and B are correct in this multiple-correct question from JEE Advanced 2020, Paper 2. Acceleration times position has the dimensions of speed squared, proving A. Acceleration times linear momentum divided by force has the dimensions of speed, proving B.

Why is option B correct in the JEE 2020 dimensional analysis question?

Option B represents acceleration times linear momentum divided by force. Its dimensions simplify as (LT^-2)(MLT^-1)/(MLT^-2) = LT^-1, which matches speed. Division by force means subtracting its exponent, not adding it.

Why are options C and D wrong in the JEE 2020 dimensional analysis question?

Option C represents acceleration times force divided by linear momentum, giving LT^-3 rather than the position dimension L. Option D multiplies acceleration, linear momentum and force, giving M^2L^3T^-5 rather than the speed dimension LT^-1. Neither is a generally valid dimensional identity.

How do I avoid sign mistakes in dimensional analysis exponent questions?

Write the physical product or quotient first, substitute dimensions or powers second, and combine exponents last. Multiplication adds exponents, while division subtracts them. Keep the fraction bar until the final step so that a denominator does not accidentally become a multiplier.