Find dimensions of each quantity in List-I

(A) Young’s Modulus

Young’s modulus = stress / strain.

Since strain is dimensionless, dimensions of Young’s modulus are same as stress.

$\text{Stress} = \frac{\text{Force}}{\text{Area}}$

Force has dimensions: $[F] = MLT^{-2}$

Area has dimensions: $[A] = L^2$

So, $[Y] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$

Hence, $(A) \to (II)$

(B) Torque

Torque = force $\times$ perpendicular distance.

$[\tau] = (MLT^{-2})(L) = ML^2T^{-2}$

Hence, $(B) \to (IV)$

(C) Coefficient of Viscosity

From Newton’s law of viscosity, $F = \eta A \frac{dv}{dx}$

Therefore, $\eta = \frac{F}{A(dv/dx)}$

Now, $[F] = MLT^{-2}, \quad [A] = L^2, \quad \left[\frac{dv}{dx}\right] = \frac{LT^{-1}}{L} = T^{-1}$

So, $[\eta] = \frac{MLT^{-2}}{L^2 \cdot T^{-1}} = ML^{-1}T^{-1}$

Hence, $(C) \to (I)$

(D) Gravitational Constant

From Newton’s law of gravitation, $F = G\frac{m_1m_2}{r^2}$

So, $G = \frac{Fr^2}{m_1m_2}$

Therefore, $[G] = \frac{(MLT^{-2})(L^2)}{M^2} = M^{-1}L^3T^{-2}$

Hence, $(D) \to (III)$

Final matching

$(A)-(II), \quad (B)-(IV), \quad (C)-(I), \quad (D)-(III)$

Compare with options

This corresponds to Option D.

Comparison with stored correct answer

Stored correct answer = D

Derived answer = D

Hence, they agree.