An alternating current in circuit is given by $I=20\sin (100\pi t+0.005\pi )A.$ The r.m.s value and the frequency of current respectively are:
$A.\quad 10A$ and $100Hz$
$B.\quad 10A$ and $50Hz$
$C.\quad 10\sqrt{2}A$ and $50Hz$
$A.\quad 10A$ and $100Hz$

Hint: One must know, how to find the RMS value of an alternating current, that is for alternating current $I={{I}_{0}}\sin \omega t$ the RMS value is\[\dfrac{I_{0}^{{}}}{\sqrt{2}}.\]The frequency of the alternating current can be found out from the angular frequency of the alternating current, that is, $\omega =\dfrac{2\pi }{T}=2\pi \vartheta .$

Step by step solution:
Let’s first start by understanding how to find out the RMS value of any function f(x). The average value of the function f(x) having time period (T) is, Average value $=\dfrac{1}{T}\int\limits_{0}^{T}{f(x)dx}$.
Therefore, the average value of square of f(x) is, $\dfrac{1}{T}\int\limits_{0}^{T}{{{[f(x)]}^{2}}dx}$.
Hence, the RMS value of of the function f(x) will be, RMS value $=\sqrt{\dfrac{1}{T}\int\limits_{0}^{T}{f(x)dx}}$.
Now, let’s consider the case for the alternating current given by $I={{I}_{0}}\sin \omega t$. Since, this is a sinusoidal wave equation, hence, the time period is $2\pi .$ Hence, using these values, that is T=$2\pi $and f(x)= $I={{I}_{0}}\sin \omega t$. Therefore, the RMS value becomes, $\sqrt{\dfrac{1}{2\pi }\int\limits_{0}^{2\pi }{{{({{I}_{0}}\sin \omega t)}^{2}}dt}}$.
That is, $\sqrt{\dfrac{I_{0}^{2}}{2\pi }\int\limits_{0}^{2\pi }{{{(\sin \omega t)}^{2}}dt}}$, here we will substitute the value of ${{\sin }^{2}}(\omega t)$ as ${{\sin }^{2}}(\omega t)=\dfrac{1}{2}(1-\cos 2\omega t).$ Further, we will also take out the constant values. This makes the RMS value to be, $\dfrac{I_{0}^{{}}}{\sqrt{2\pi }}\sqrt{\int\limits_{0}^{2\pi }{\dfrac{1}{2}(1-\cos 2\omega t)dt}}$. That is, $\dfrac{I_{0}^{{}}}{2\sqrt{\pi }}\sqrt{\int\limits_{0}^{2\pi }{(1-\cos 2\omega t)dt}}$.
We know that,$\omega =\dfrac{2\pi }{T}$. Hence the integral becomes, $\dfrac{I_{0}^{{}}}{2\sqrt{\pi }}\sqrt{[t]_{0}^{2\pi }-[\dfrac{\sin \omega t}{\omega }]_{0}^{2\pi }}=\dfrac{I_{0}^{{}}}{2\sqrt{\pi }}\sqrt{[2\pi -0]-[\dfrac{\sin 2\pi -\sin 0}{\omega }]}$.
Therefore, the RMS eventually becomes, \[\dfrac{I_{0}^{{}}}{2\sqrt{\pi }}\sqrt{2\pi }=\dfrac{I_{0}^{{}}}{\sqrt{2}}.\]
From the given value of alternating current in the problem, $I=20\sin (100\pi t+0.005\pi )A$, we get${{I}_{0}}=20$. Hence, the RMS value of the alternating current is, \[\dfrac{I_{0}^{{}}}{\sqrt{2}}=\dfrac{20}{\sqrt{2}}=10\sqrt{2}A.\]
Simultaneously, the angular frequency of the alternating current value is $\omega =\dfrac{2\pi }{T}=2\pi \vartheta ,$where$\vartheta $ is the frequency of the alternating current.
In the given problem, $\omega =100\pi $. Equating it with $\omega =\dfrac{2\pi }{T}=2\pi \vartheta ,$we get, $\omega =100\pi =2\pi \vartheta \Rightarrow \vartheta =50Hz.$ That is the frequency of the alternating current is 50Hz.

Note: You may think that we are always considering the value of alternating current as $I={{I}_{0}}\sin \omega t$ for most of the equations. The reason is, that this value of ${{I}_{0}}\sin \omega t$ is the only necessary part of the alternating current carrying the important information required for the calculations done above. The additional term is the phase angle.
When we consider, ${{I}_{1}}={{I}_{0}}\sin (\omega t\pm \delta )$, here $(\delta )$ refers to the phase angle. The phase angle is the amount or angle with which the alternating current$({{I}_{1}})$ will lead or lag against $I={{I}_{0}}\sin \omega t$.
$({{I}_{1}})$will be leading against$I={{I}_{0}}\sin \omega t$, when the phase angle is positive. That is, ${{I}_{1}}={{I}_{0}}\sin (\omega t+\delta )$. Similarly, $({{I}_{1}})$ will be lagging behind$I={{I}_{0}}\sin \omega t$, when the phase angle is negative. That is, ${{I}_{1}}={{I}_{0}}\sin (\omega t-\delta )$.