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JEE Advanced 2013 · Units and Measurements

Match List I with List II and select the correct answer using the codes given below the lists: | List I | List II | | --- | --- | | PP. Boltzmann constant | 1. [ML2T−1][ML^{2}T^{-1}] | | QQ. Coefficient of viscosity | 2. [ML−1T−1][ML^{-1}T^{-1}] | | RR. Planck constant | 3. [MLT−3K−1][MLT^{-3}K^{-1}] | | SS. Thermal conductivity | 4. [ML2T−2K−1][ML^{2}T^{-2}K^{-1}] | PQRSPQRS is

Answer: C. 42134 2 1 3

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Written solution

Step 1: Find the dimensional formula for Boltzmann constant (P).

The Boltzmann constant kBk_B relates energy and temperature.

kB=EnergyTemperaturek_B = \frac{\text{Energy}}{\text{Temperature}}

Substitute the dimensions of energy [ML2T−2][ML^{2}T^{-2}] and temperature [K][K].

[kB]=[ML2T−2][K]=[ML2T−2K−1][k_B] = \frac{[ML^{2}T^{-2}]}{[K]} = [ML^{2}T^{-2}K^{-1}]

This matches with entry 4 in List II.

P→4P \to 4

Step 2: Find the dimensional formula for the coefficient of viscosity (Q).

From Newton's law of viscosity, the viscous force is given by:

F=ηAdvdxF = \eta A \frac{dv}{dx} η=FA(dvdx)\eta = \frac{F}{A \left(\frac{dv}{dx}\right)}

Substitute the dimensions of force [MLT−2][MLT^{-2}], area [L2][L^{2}], and velocity gradient [T−1][T^{-1}].

[η]=[MLT−2][L2][T−1]=[ML−1T−1][\eta] = \frac{[MLT^{-2}]}{[L^{2}][T^{-1}]} = [ML^{-1}T^{-1}]

This matches with entry 2 in List II.

Q→2Q \to 2

Step 3: Find the dimensional formula for Planck constant (R).

Using the energy of a photon:

E=hνE = h\nu h=Eνh = \frac{E}{\nu}

Substitute the dimensions of energy [ML2T−2][ML^{2}T^{-2}] and frequency [T−1][T^{-1}].

[h]=[ML2T−2][T−1]=[ML2T−1][h] = \frac{[ML^{2}T^{-2}]}{[T^{-1}]} = [ML^{2}T^{-1}]

This matches with entry 1 in List II.

R→1R \to 1

Step 4: Find the dimensional formula for thermal conductivity (S).

From Fourier's law of heat conduction, the rate of heat flow is:

dQdt=κAdTdx\frac{dQ}{dt} = \kappa A \frac{dT}{dx} κ=dQdtA(dTdx)\kappa = \frac{\frac{dQ}{dt}}{A \left(\frac{dT}{dx}\right)}

Substitute the dimensions of heat flow rate (power) [ML2T−3][ML^{2}T^{-3}], area [L2][L^{2}], and temperature gradient [KL−1][K L^{-1}].

[κ]=[ML2T−3][L2][KL−1]=[MLT−3K−1][\kappa] = \frac{[ML^{2}T^{-3}]}{[L^{2}][K L^{-1}]} = [MLT^{-3}K^{-1}]

This matches with entry 3 in List II.

S→3S \to 3

Step 5: Combine the matches to find the code for PQRS.

Combining the individual results:

P→4,Q→2,R→1,S→3P \to 4, \quad Q \to 2, \quad R \to 1, \quad S \to 3

The correct sequence for PQRSPQRS is 4,2,1,34, 2, 1, 3.

Answer: 4,2,1,34, 2, 1, 3 (Option 3)

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