Impedance of circuit

$$\begin{aligned} & \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}\right)^2} \\ & =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \\ & =\sqrt{(100 \pi)^2+(100 \pi)^2} \\ & =\sqrt{2} \times 100 \pi \\ & \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \\ & \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere } \end{aligned}$$

Correct Answer : 1