Match List-I with List-II.

	List-I		List-II
A	Young’s Modulus	I	${\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}$
B	Torque	II	${\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}$
C	Coefficient of Viscosity	III	${\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}$
D	Gravitational Constant	IV	${\mathrm{ML}}^2{\mathrm{T}}^{-2}$

Choose the correct answer from the options given below :

Let's determine the dimensional formulas for each physical quantity in List-I and match them with List-II.

Step 1: Finding Dimensional Formulas

A) Young’s Modulus (Y)

Young’s modulus is defined as:

$$Y = \frac{\text{Stress}}{\text{Strain}}$$

• Stress = Force / Area = $\frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$
• Strain is dimensionless.

Thus, the dimensional formula of Young’s modulus is: $$ML^{-1}T^{-2}$$

➡ Matches with II.

B) Torque (τ)

Torque is defined as:

$$\tau = \text{Force} \times \text{Perpendicular distance}$$

• Force = $MLT^{-2}$
• Distance = $L$

Thus, the dimensional formula of Torque is: $$ML^2T^{-2}$$

➡ Matches with IV.

C) Coefficient of Viscosity (η)

The coefficient of viscosity is given by:

$$\eta = \frac{\text{Shear Stress}}{\text{Shear Rate}}$$

• Shear Stress = $\frac{\text{Force}}{\text{Area}} = ML^{-1}T^{-2}$
• Shear Rate = $\frac{\text{Velocity}}{\text{Distance}} = T^{-1}$

Thus, the dimensional formula of viscosity is: $$ML^{-1}T^{-1}$$

➡ Matches with I.

D) Gravitational Constant (G)

From Newton’s Law of Gravitation:

$$F = G \frac{m_1 m_2}{r^2}$$

Rearranging for $G$:

$$G = \frac{F r^2}{m_1 m_2}$$

Substituting Force ($F = MLT^{-2}$) and mass ($M$):

$$G = \frac{MLT^{-2} \times L^2}{M \times M}$$ $$= M^{-1}L^3T^{-2}$$

➡ Matches with III.

Step 2: Matching List-I with List-II

Final Answer:

$$\mathbf{A \to II, \quad B \to IV, \quad C \to I, \quad D \to III}$$