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: \[ \boldsymbol{0} \]

Answer 2

Step 1: Understand the given question.
We are asked to find the dimensional formula of a quantity which represents modulus of elasticity per unit torque. It is given that the dimensions are expressed as \( [M^a L^b T^c] \), and we are told that \( b = 3 \). We need to find the value of \( c \).

Step 2: Write the dimensional formula for each physical quantity.
- Modulus of elasticity (or stress/strain) has the same dimensions as pressure:
\[ [\text{Modulus of elasticity}] = [M^1 L^{-1} T^{-2}] \] - Torque = Force × Distance
\[ [\text{Torque}] = [M^1 L^2 T^{-2}] \]

Step 3: Compute the dimensional formula for "modulus of elasticity per unit torque".
\[ \frac{\text{Modulus of elasticity}}{\text{Torque}} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1-1} L^{-1-2} T^{-2+2}] = [M^0 L^{-3} T^0] \]

Step 4: Compare with the given form.
The general dimensional formula is \( [M^a L^b T^c] \). Comparing terms, we have:
\[ a = 0, \quad b = -3, \quad c = 0 \] But since the problem states \( b = 3 \) (considering magnitude only), we find:
\[ c = 0 \]

Final Answer:
\[ \boxed{c = 0} \]