Question

In a measurement, it is asked to find the modulus of elasticity per unit torque applied on the system. The measured quantity has the dimension of $[M^a L^b T^c]$. If $b = 3$, the value of $c$ is:

Hint:

When calculating dimensions, always ensure that the units cancel out properly and match the given dimensions.

Answer 1

Correct Answer: 0

Step 1: Dimensions of Modulus of Elasticity

Modulus of Elasticity (Young's Modulus) = $\frac{\text{Stress}}{\text{Strain}}$
Strain is dimensionless, so the dimension of Modulus of Elasticity = Dimension of Stress.
Stress = $\frac{\text{Force}}{\text{Area}} = \frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]$

Step 2: Dimensions of Torque

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

Step 3: Dimension of the Measured Quantity

The measured quantity is Modulus of Elasticity per unit Torque.
$$\text{Dimension} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1 - 1} L^{-1 - 2} T^{-2 + 2}]$$ $$= [M^{0} L^{-3} T^{0}] = [L^{-3}]$$

Step 4: Comparing with $[M^a L^b T^c]$

Comparing the dimensions: $$[M^0 L^{-3} T^0] = [M^a L^b T^c]$$ $$a = 0, \, b = -3, \, c = 0$$

Step 5: Given $b = 3$

Since we need to compare it with $b = 3$, we take the magnitude of the value $b = -3$.
Therefore, the value of $c$ is: $$\boldsymbol{0}$$