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MCQEasyJEE 2024Dimensions & Dimensional Analysis

JEE Physics 2024 Question with Solution

A force is represented by F=ax2+bt1/2F = ax^2 + bt^{1/2} where xx = distance and tt = time. The dimensions of b2a\frac{b^2}{a} are:

  • A

    [ML3T−3][ML^3T^{-3}]

  • B

    [MLT−2][MLT^{-2}]

  • C

    [ML2T−1][ML^2T^{-1}]

  • D

    [MLT2][MLT^2]

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: F=ax2+bt1/2F = ax^2 + bt^{1/2}, where [F]=[MLT−2][F] = [MLT^{-2}], [x]=[L][x] = [L] and [t]=[T][t] = [T].

Find: The dimensions of b2a\frac{b^2}{a}.

Since each term in the equation must have the dimensions of force:

[a][L2]=[MLT−2][a][L^2] = [MLT^{-2}]

So,

[a]=[ML−1T−2][a] = [ML^{-1}T^{-2}]

For the second term,

[b][T1/2]=[MLT−2][b][T^{1/2}] = [MLT^{-2}]

Hence,

[b]=[MLT−5/2][b] = [MLT^{-5/2}]

Now,

[b2a]=[b]2[a]=[M2L2T−5][ML−1T−2]\left[\frac{b^2}{a}\right] = \frac{[b]^2}{[a]} = \frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]}

Therefore,

[b2a]=[ML3T−3]\left[\frac{b^2}{a}\right] = [ML^3T^{-3}]

Thus, the dimensions of b2a\frac{b^2}{a} are [ML3T−3][ML^3T^{-3}]. The correct option is A.

Using dimensional consistency of each term

Given: F=ax2+bt1/2F = ax^2 + bt^{1/2}.

Find: The dimensions of b2a\frac{b^2}{a}.

The force equation contains two terms added together, so both ax2ax^2 and bt1/2bt^{1/2} must separately have the dimensions of force.

For ax2ax^2:

[a][x2]=[F][a][x^2] = [F] [a][L2]=[MLT−2][a][L^2] = [MLT^{-2}] [a]=[ML−1T−2][a] = [ML^{-1}T^{-2}]

For bt1/2bt^{1/2}:

[b][T1/2]=[MLT−2][b][T^{1/2}] = [MLT^{-2}] [b]=[MLT−5/2][b] = [MLT^{-5/2}]

Now square the dimensions of bb:

[b]2=[M2L2T−5][b]^2 = [M^2L^2T^{-5}]

Then divide by [a][a]:

[b]2[a]=[M2L2T−5][ML−1T−2]\frac{[b]^2}{[a]} = \frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]}

Subtract powers in division:

[M2−1L2−(−1)T−5−(−2)]=[ML3T−3][M^{2-1}L^{2-(-1)}T^{-5-(-2)}] = [ML^3T^{-3}]

Therefore, the required dimensions are [ML3T−3][ML^3T^{-3}] and the correct option is A.

Common mistakes

  • A common mistake is taking [b][T1/2]=[MLT−2][b][T^{1/2}] = [MLT^{-2}] and writing [b]=[MLT−3/2][b] = [MLT^{-3/2}]. This is wrong because dividing by T1/2T^{1/2} means the exponent of TT becomes −2−12=−52-2 - \frac{1}{2} = -\frac{5}{2}. Use [b]=[MLT−5/2][b] = [MLT^{-5/2}] instead.

  • Another mistake is forgetting that terms added in an equation must have the same dimensions. In F=ax2+bt1/2F = ax^2 + bt^{1/2}, both terms must individually match the dimensions of force. Do not compare only the whole right-hand side with FF without splitting the terms.

  • Students also make errors while dividing dimensional formulas, especially with negative powers. In [M2L2T−5][ML−1T−2]\frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]}, the power of LL becomes 2−(−1)=32 - (-1) = 3 and the power of TT becomes −5−(−2)=−3-5 - (-2) = -3. Carefully subtract exponents while dividing.

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