JEE Physics 2025 Question with Solution

Given: The measured quantity is modulus of elasticity per unit torque and its dimensions are written as $[M^a L^b T^c]$.

Find: The value of $c$.

Dimensions of modulus of elasticity:

$$\text{Modulus of Elasticity} = \frac{\text{Stress}}{\text{Strain}}$$

Since strain is dimensionless,

$$[\text{Modulus of Elasticity}] = [\text{Stress}] = \frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]$$

Dimensions of torque:

$$[\text{Torque}] = [\text{Force}] \times [\text{Distance}] = [M^1 L^1 T^{-2}] \times [L] = [M^1 L^2 T^{-2}]$$

Measured quantity:

$$\frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1-1} L^{-1-2} T^{-2+2}] = [M^0 L^{-3} T^0]$$

Comparing with $[M^a L^b T^c]$,

$$a = 0, \quad b = -3, \quad c = 0$$

the solution notes the given $b = 3$ by magnitude, but the time exponent remains unchanged.

Therefore, the value of $c$ is $0$.

Given: The quantity is modulus of elasticity per unit torque.

Find: The exponent of time, $c$.

Modulus of elasticity has the same dimensions as pressure:

$$[M^1 L^{-1} T^{-2}]$$

Torque has dimensions:

$$[M^1 L^2 T^{-2}]$$

So the required dimensional formula is:

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

Subtract the exponents while dividing:

$$M^{1-1} = M^0, \quad L^{-1-2} = L^{-3}, \quad T^{-2-(-2)} = T^0$$

Hence,

$$[M^a L^b T^c] = [M^0 L^{-3} T^0]$$

So, the exponent of $T$ is $0$.

Therefore, the numerical answer is $0$.