Calculate the inductance of a coil of resistance 30 ohms connected to a 100volt A.c source if the coil draw an Ir.m.s of 2A

The impedance is the total opposition to current flow in an AC circuit and can be found using Ohm's law for AC circuits:

$$Z = \frac{V_{rms}}{I_{rms}}$$ $$Z = \frac{100\, \text{V}}{2\, \text{A}}$$ $$Z = 50\, \Omega$$

For a coil with resistance, the impedance is related to the resistance and inductive reactance by the formula:

$$Z = \sqrt{R^2 + X_L^2}$$

We can rearrange this to solve for $X_L$:

$$X_L^2 = Z^2 - R^2$$ $$X_L = \sqrt{Z^2 - R^2}$$

Substitute the values of $Z$ and $R$:

$$X_L = \sqrt{(50\, \Omega)^2 - (30\, \Omega)^2}$$ $$X_L = \sqrt{2500\, \Omega^2 - 900\, \Omega^2}$$ $$X_L = \sqrt{1600\, \Omega^2}$$ $$X_L = 40\, \Omega$$

The inductive reactance is also related to the inductance ($L$) and the angular frequency ($\omega$) by the formula:

$$X_L = \omega L$$

Where $\omega = 2\pi f$, and $f$ is the frequency of the AC source.
The problem does not provide the frequency of the AC source. Therefore, we will express the inductance in terms of the angular frequency $\omega$.

$$L = \frac{X_L}{\omega}$$ $$L = \frac{40\, \Omega}{\omega}$$

If we assume a standard AC frequency, for example, $f = 50\, \text{Hz}$ (a common frequency in many regions):
The angular frequency would be:

$$\omega = 2\pi f = 2\pi (50\, \text{Hz}) = 100\pi\, \text{rad/s}$$

Now, substitute this value into the equation for $L$:

$$L = \frac{40\, \Omega}{100\pi\, \text{rad/s}}$$ $$L = \frac{2}{5\pi}\, \text{H}$$ $$L \approx \frac{2}{5 \times 3.14159}$$ $$L \approx \frac{2}{15.70795}$$ $$L \approx 0.1273\, \text{H}$$

The inductance of the coil is $\boxed{\frac{40}{\omega}\, \text{H}}$.
Assuming a frequency of $50\, \text{Hz}$, the inductance is approximately $\boxed{0.127\, \text{H}}$.