MCQMediumJEE 2025Dimensions & Dimensional Analysis

JEE Physics 2025 Question with Solution

Match List-I with List-II:

A matching table between List-I and List-II showing physical quantities and their dimensional formulas or units to be matched.

Choose the correct answer from the options given below:

  • A

    (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

  • B

    (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

  • C

    (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

  • D

    (A)-(III), (B)-(III), (C)-(IV), (D)-(I)

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A match-the-following question on dimensional analysis.

Find: The correct option for matching the physical quantities with their dimensional formulas.

From the solution:

  • Young's Modulus is stress/strain, so its dimension is
MLT2L2=ML1T2\frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}
  • Torque is force ×\times distance, so its dimension is
MLT2×L=ML2T2MLT^{-2} \times L = ML^2T^{-2}
  • Coefficient of Viscosity is obtained from
η=Force/AreaVelocity gradient=ML1T2T1=ML1T1\eta = \frac{\text{Force/Area}}{\text{Velocity gradient}} = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}
  • Gravitational Constant from Newton's law has dimension
G=Fr2m1m2=(MLT2)L2M2=M1L3T2G = \frac{Fr^2}{m_1 m_2} = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}

the solution explicitly states: The Correct Option is D.

It also notes the final matching as

(A)(I),  (B)(II),  (C)(I),  (D)(IV)(A)-(I), \; (B)-(II), \; (C)-(I), \; (D)-(IV)

which does not exactly match the listed options. Since the solution explicitly declares option D as correct, the answer is taken as D in accordance with the solution.

Therefore, the correct option is D.

Detailed Extraction with Discrepancy Note

Given: The quantities are to be matched using their dimensional formulas.

Find: Which option among A, B, C, D is marked correct by the solution.

The extracted working shows:

  1. Young's Modulus
Young’s Modulus=StressStrain\text{Young's Modulus} = \frac{\text{Stress}}{\text{Strain}}

Since strain is dimensionless and stress is force per area,

Stress=MLT2L2=ML1T2\text{Stress} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}

So Young's Modulus corresponds to ML1T2ML^{-1}T^{-2}.

  1. Torque
Torque=Force×distance\text{Torque} = \text{Force} \times \text{distance}

Hence,

MLT2×L=ML2T2MLT^{-2} \times L = ML^2T^{-2}

So Torque corresponds to ML2T2ML^2T^{-2}.

  1. Coefficient of Viscosity Using
ηdvdy=ForceArea\eta \frac{dv}{dy} = \frac{\text{Force}}{\text{Area}}

we get

η=ML1T2T1=ML1T1\eta = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}

So the coefficient of viscosity corresponds to ML1T1ML^{-1}T^{-1}.

  1. Gravitational Constant From
F=Gm1m2r2F = G\frac{m_1m_2}{r^2}

we obtain

G=Fr2m1m2=(MLT2)L2M2=M1L3T2G = \frac{Fr^2}{m_1m_2} = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}

So gravitational constant corresponds to M1L3T2M^{-1}L^3T^{-2}.

The solution text contains an internal inconsistency: the dimensional matching written in the steps does not align cleanly with the listed options, and one portion even says "Thus, the correct answer is option (4)." However, the solution clearly states The Correct Option is D.

Therefore, using the solution, the correct option is D.

Common mistakes

  • Students often confuse stress with force while finding Young's modulus. This is wrong because Young's modulus is stress/strain, not force/strain. First convert stress to force per unit area, then use strain as dimensionless.

  • A common mistake is to treat torque as having the same dimensions as force. This is incorrect because torque includes an extra factor of distance. Always use Torque=Force×distance\text{Torque} = \text{Force} \times \text{distance}.

  • For coefficient of viscosity, students may forget the velocity gradient term and use only force per area. This gives the wrong dimensions. Use η=Force/Areadv/dy\eta = \frac{\text{Force/Area}}{dv/dy}, and note that dv/dydv/dy has dimension T1T^{-1}.

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