Question

Match List-I with List-II. 

List-I  (A) Coefficient of viscosity  (B) Intensity of wave  (C) Pressure gradient  (D) Compressibility

List-II  (I) [ML-1T-1]  (II) [MT-3]  (III) [ML-2T-2]  (IV) [M-1LT2]

• A. (A)–(I), (B)–(IV), (C)–(III), (D)–(I)
• B. (A)–(I), (B)–(II), (C)–(III), (D)–(IV)
• C. (A)–(IV), (B)–(II), (C)–(III), (D)–(I)
• D. (A)–(IV), (B)–(I), (C)–(II), (D)–(III)

Hint:

In dimensional analysis, the dimensions of physical quantities are crucial for understanding their relationships. Pay attention to how exponents are used to represent various physical properties.

Answer 1

The Correct Option is B

The problem requires us to find the dimensional formulas for the physical quantities in List-I and match them with the correct dimensions provided in List-II.

Concept Used:

Dimensional analysis involves expressing physical quantities in terms of fundamental dimensions: Mass (M), Length (L), and Time (T). To find the dimension of a quantity, we use its definition or a formula relating it to other quantities with known dimensions.

Key dimensional formulas used in derivations:

• Force \( [F] = [mass \times acceleration] = [MLT^{-2}] \)
• Area \( [A] = [L^2] \)
• Pressure \( [P] = \frac{[F]}{[A]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] \)
• Energy/Work \( [E] = [F \times distance] = [ML^2T^{-2}] \)
• Power \( [\text{Power}] = \frac{[E]}{[T]} = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}] \)

Step-by-Step Solution:

Step 1: (A) Coefficient of viscosity (\( \eta \))

According to Newton's law of viscosity, the viscous force is given by \( F = \eta A \frac{dv}{dx} \). We can rearrange this to find the dimensions of \( \eta \).

\[ [\eta] = \frac{[F]}{[A] \cdot [\frac{dv}{dx}]} = \frac{[MLT^{-2}]}{[L^2] \cdot \frac{[LT^{-1}]}{[L]}} = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} \] \[ [\eta] = [ML^{-1}T^{-1}] \]

This matches with (I) in List-II.

Step 2: (B) Intensity of wave (I)

The intensity of a wave is defined as the power transmitted per unit area.

\[ [I] = \frac{[\text{Power}]}{[\text{Area}]} = \frac{[ML^2T^{-3}]}{[L^2]} \] \[ [I] = [MT^{-3}] \]

This matches with (II) in List-II.

Step 3: (C) Pressure gradient

Pressure gradient is the rate of change of pressure with respect to distance.

\[ [\text{Pressure Gradient}] = \frac{[\text{Pressure}]}{[\text{Distance}]} = \frac{[ML^{-1}T^{-2}]}{[L]} \] \[ [\text{Pressure Gradient}] = [ML^{-2}T^{-2}] \]

This matches with (III) in List-II.

Step 4: (D) Compressibility (K)

Compressibility is the reciprocal of the Bulk Modulus (B). The Bulk Modulus has the same dimensions as pressure.

\[ [K] = \frac{1}{[\text{Bulk Modulus}]} = \frac{1}{[\text{Pressure}]} \] \[ [K] = \frac{1}{[ML^{-1}T^{-2}]} = [M^{-1}L^{1}T^{2}] \]

This matches with (IV) in List-II.

Final Result:

Based on the derivations, the correct matching is:

• (A) Coefficient of viscosity → (I) \( [ML^{-1}T^{-1}] \)
• (B) Intensity of wave → (II) \( [MT^{-3}] \)
• (C) Pressure gradient → (III) \( [ML^{-2}T^{-2}] \)
• (D) Compressibility → (IV) \( [M^{-1}LT^{2}] \)

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