JEE Main 2025 Jan 29 Shift 2, Physics Q16: Units And Dimensions

JEE Main2025Jan 29, Shift 2Physics

Match List-I with List-II.

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

Choose the correct answer from the options given below :

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

Solution

Derive each dimensional formula.

(A) Young's modulus: $Y = \frac{Stress}{Strain} = \frac{F / A}{Δ L / L}$ , dimensions $\frac{M L T ^{- 2}}{L ^{2}} = M L^{- 1} T^{- 2}$ . $\to$ (II).

(B) Torque: $τ = r \times F$ , dimensions $L \cdot M L T^{- 2} = M L^{2} T^{- 2}$ . $\to$ (IV).

(C) Coefficient of viscosity: $F = η A (d v / d y)$ , so $[η] = \frac{M L T ^{- 2}}{L ^{2} \cdot T ^{- 1}} = M L^{- 1} T^{- 1}$ . $\to$ (I).

(D) Gravitational constant: $F = GM m / r^{2}$ , so $[G] = \frac{M L T ^{- 2} \cdot L ^{2}}{M ^{2}} = M^{- 1} L^{3} T^{- 2}$ . $\to$ (III).

$∴$ The matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

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JEE Main 2025 Jan 29 Shift 2, Physics Q16: Units And Dimensions

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