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: Derivation of Modulus of Elasticity Dimensions

Modulus of Elasticity (Young's Modulus) is defined as \( \frac{\text{Stress}}{\text{Strain}} \).
As strain is dimensionless, the dimensions of Modulus of Elasticity are equivalent to the dimensions of Stress.
Stress is calculated as \( \frac{\text{Force}}{\text{Area}} \), which translates to \( \frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}] \).

Step 2: Derivation of Torque Dimensions

Torque is calculated as \( \text{Force} \times \text{Distance} \).
The dimensions are \( [M^1 L^1 T^{-2}] \times [L] = [M^1 L^2 T^{-2}] \).

Step 3: Determining the Dimension of the Measured Quantity

The measured quantity is the ratio of Modulus of Elasticity to Torque.
The dimensional analysis is as follows: \[ \text{Dimension} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1 - 1} L^{-1 - 2} T^{-2 + 2}] \] \[ = [M^{0} L^{-3} T^{0}] = [L^{-3}] \]

Step 4: Comparison with the General Form \( [M^a L^b T^c] \)

Comparing the derived dimensions with the general form: \[ [M^0 L^{-3} T^0] = [M^a L^b T^c] \] This yields the exponents: \[ a = 0, \, b = -3, \, c = 0 \]

Step 5: Given \( b = 3 \) and Determining \( c \)

Given the requirement to compare with \( b = 3 \), we consider the magnitude of the calculated exponent for length, which is \( |b| = |-3| = 3 \).
The value of \( c \) is therefore: \[ \boldsymbol{0} \]