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

We need to find the dimension of the ratio $$\frac{\text{Modulus of Elasticity}}{\text{Torque}}$$ and determine the value of $$c$$ given that $$b = -3$$.

To begin, we write the dimension of Modulus of Elasticity ($$E$$). Modulus of elasticity (Young's modulus) is defined as stress/strain. Since strain is dimensionless:

$$ [E] = [\text{Stress}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] $$

Next, we write the dimension of Torque ($$\tau$$). Torque is given by Force $$\times$$ distance (lever arm):

$$ [\tau] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}] $$

We then compute the dimension of $$E/\tau$$:

$$ \left[\frac{E}{\tau}\right] = \frac{[M L^{-1} T^{-2}]}{[M L^2 T^{-2}]} $$

Dividing each base dimension:

$$ M: M^1 / M^1 = M^0 $$

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

$$ T: T^{-2} / T^{-2} = T^0 $$

Therefore,

$$ \left[\frac{E}{\tau}\right] = [M^0 L^{-3} T^0] $$

Comparing this result with $$[M^a L^b T^c]$$ shows that $$a = 0$$, $$b = -3$$, and $$c = 0$$. This confirms the given value $$b = -3$$ and yields the value of $$c = 0$$.

The answer is 0.