Question

Match List-I with List-II.

Choose the correct answer from the options given below:

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

Hint:

To match physical quantities with their units, use dimensional analysis. Dimensional analysis helps in determining the correct formula and unit for each physical quantity.

Answer 1

The Correct Option is D

(A) Young's Modulus: Young's modulus is the ratio of stress to strain. Its SI unit is \( \frac{\text{Force}}{\text{Area}} = \frac{MLT^{-2}}{L^2} = M L T^{-2} \), which corresponds to (I).

(B) Torque: Torque is the product of force and distance. The SI unit of force is \( M L T^{-2} \), and the unit of distance is \( L \), so the SI unit of torque is \( M L^2 T^{-2} \), which corresponds to (II).

(C) Coefficient of Viscosity: The coefficient of viscosity is the ratio of force per unit area to the rate of change of velocity. Its unit is \( M L^{-1} T^{-1} \), which corresponds to (I).

(D) Gravitational Constant: The gravitational constant has the unit \( M^{-1} L^3 T^{-2} \), which corresponds to (IV). Thus, the correct answer is option (4).

Answer 2

Step 1: Recall the dimensional formulas of physical quantities.
We are to match the given quantities with their respective dimensional formulas.

(A) Young’s Modulus:
Young’s Modulus = Stress / Strain.
Stress has the dimension of pressure = Force / Area = \( \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \).
Strain is dimensionless.
Hence, the dimensional formula of Young’s Modulus = \( ML^{-1}T^{-2} \).
So, (A) → (I).

(B) Torque:
Torque = Force × Perpendicular distance.
Force = \( MLT^{-2} \), Distance = \( L \).
Hence, Torque = \( ML^2T^{-2} \).
So, (B) → (II).

(C) Coefficient of Viscosity:
Viscous force per unit area = \( \eta \frac{dv}{dy} \).
Thus, \( \eta = \frac{\text{Force/Area}}{\text{Velocity gradient}} = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1} \).
Hence, (C) → (I).

(D) Gravitational Constant:
From Newton’s law of gravitation:
\[ F = G \frac{m_1 m_2}{r^2} \] \[ G = \frac{Fr^2}{m_1 m_2} = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}. \] Hence, (D) → (IV).

Step 2: Final matching.
(A) – (I)
(B) – (II)
(C) – (I)
(D) – (IV)

Final Answer:
\[ \boxed{(A)-(I), \, (B)-(II), \, (C)-(I), \, (D)-(IV)} \]