Question

Match column I with column II and choose the correct option.

Column I	Column II
(I) Torque	(a) M°LT-2
(II) stress	(b) ML-1T-1
(III) Coefficient of viscosity	(c) ML-IT-2
(IV) Potential gradient	(d) ML27-2

• A. I → a, II → c, III → b, IV → d
• B. I → d, II → b, III → c, IV → a
• C. I → d, II → c, III → b, IV → a
• D. I → a, II → c, III → d, IV → b

Answer 1

The Correct Option is C

(A) Torque: Torque (\(T\)) is given by the product of force (\(F\)) and distance (\(L\)): \[ [\text{Torque}] = F \cdot L. \] The dimensional formula for force is: \[ [F] = MLT^{-2}. \] Thus, the dimensional formula for torque is: \[ [\text{Torque}] = MLT^{-2} \cdot L = ML^2T^{-2}. \] So, \(A \to I\). (B) Stress: Stress (\(\sigma\)) is defined as force (\(F\)) per unit area (\(A\)): \[ [\text{Stress}] = \frac{F}{A}. \] The dimensional formula for area is: \[ [A] = L^2. \] Therefore: \[ [\text{Stress}] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}. \] So, \(B \to IV\). (C) Pressure Gradient: Pressure gradient is defined as the change in pressure per unit length (\(\Delta P / \Delta L\)): \[ \left[\frac{\Delta P}{\Delta L}\right] = \frac{F/A}{L}. \] Substituting the dimensional formula for pressure (\(F/A\)): \[ [\text{Pressure Gradient}] = \frac{ML^{-1}T^{-2}}{L} = ML^{-2}T^{-2}. \] So, \(C \to I\). (D) Coefficient of Viscosity: The coefficient of viscosity (\(\eta\)) is defined as: \[ [\eta] = \frac{\text{Force per unit area}}{\text{Velocity gradient}}. \] The dimensional formula for velocity gradient is: \[ \left[\frac{v}{L}\right] = T^{-1}. \] Therefore: \[ [\eta] = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}. \] So, \(D \to III\). Final Answer: The correct match is: \[ \boxed{\text{A-II, B-IV, C-I, D-III}}. \]