JEE Main 2025 — 28 January, Morning Shift — Question 69

Dimensions of Modulus of Elasticity ($E$):

Modulus of elasticity is defined as $\frac{\text{Stress}}{\text{Strain}}$. Since strain is dimensionless, its dimensions are equal to those of stress (Force per unit area):

$$[E] = \frac{[M L T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]$$

Dimensions of Torque ($\tau$):

Torque is the product of force and distance:

$$[\tau] = [M L T^{-2}] \times [L] = [M^1 L^2 T^{-2}]$$

Dimensions of Modulus of Elasticity per unit Torque:

$$\frac{E}{\tau} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1-1} L^{-1-2} T^{-2 - (-2)}] = [M^0 L^{-3} T^0]$$

Comparing this result $[M^0 L^{-3} T^0]$ with the general form $\left[M^a L^b T^c\right]$, we get:

$a = 0$

$b = -3$

$c = 0$

Taking the magnitude for $b$ as given ($b = 3$), the corresponding power for time remains $0$.

Thus, the value of $c$ is 0.