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

Modulus of Elasticity (Young’s Modulus, $E$)

• Dimensional formula: $$E = \frac{\text{Stress}}{\text{Strain}}$$
• Stress = Force / Area = $\frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2}$
• Strain is dimensionless.
• So, dimensional formula of $E$: $$[E] = M L^{-1} T^{-2}$$

Torque ($\tau$)

• Torque = Force × Distance
• $\tau = (M L T^{-2}) \times L = M L^2 T^{-2}$
• So, dimensional formula of $\tau$: $$[\tau] = M L^2 T^{-2}$$

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

Canceling $M$ and $T^{-2}$:

$$= L^{-1 - 2} = L^{-3}$$

Thus, the dimensional formula of the required quantity is:

$$[M^0 L^{-3} T^0]$$

Compare with $M^a L^b T^c$

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