JEE Physics 2025 Question with Solution

Given: Match the quantities in LIST-I with the dimensional formulae in LIST-II.

Find: The correct option for the matching.

For the Boltzmann constant $k$, use the ideal gas law:

Here, pressure has dimensions $ML^{-1}T^{-2}$, volume has dimensions $L^3$, $N$ is dimensionless, and temperature has dimension $K$.

So,

Thus, A matches with III.

For the coefficient of viscosity $\eta$, use the viscous force relation:

where force has dimensions $MLT^{-2}$, radius has dimensions $L$, and velocity has dimensions $LT^{-1}$.

Hence,

Thus, B matches with IV.

For Planck's constant $h$, use:

Energy has dimensions $ML^2T^{-2}$ and frequency has dimensions $T^{-1}$.

Therefore,

Thus, C matches with I.

For thermal conductivity $K$, use the heat-flow equation:

Here, $\frac{dQ}{dt}$ is power with dimensions $ML^2T^{-3}$, area has dimensions $L^2$, and temperature gradient has dimensions $KL^{-1}$.

So,

Thus, D matches with II.

Hence the correct matching is A-III, B-IV, C-I, D-II. Therefore, the correct option is A.

Given: Four physical quantities and four dimensional formulae.

Find: Which option gives the correct one-to-one correspondence.

Using known dimensional formulae directly:

• Boltzmann constant $[k] = ML^2T^{-2}K^{-1}$
• Coefficient of viscosity $[\eta] = ML^{-1}T^{-1}$
• Planck's constant $[h] = ML^2T^{-1}$
• Thermal conductivity $[\kappa] = MLT^{-3}K^{-1}$

Now compare with LIST-II:

• III is $ML^2T^{-2}K^{-1}$
• IV is $ML^{-1}T^{-1}$
• I is $ML^2T^{-1}$
• II is $MLT^{-3}K^{-1}$

Therefore:

• A $\to$ III
• B $\to$ IV
• C $\to$ I
• D $\to$ II

So the correct option is A.