The mutual induction between primary and secondary coils of a transformer is $2H$. If an AC current of peak value $5{\text{A}}$ and frequency $50Hz$ flows through the primary coil, then the peak voltage in secondary coil is
(A) $10\pi $ volts
(B) $10$ volts
(C) $1000\pi $ volts
(D) $20$ volts

Hint: To solve this question, we need to find out the flux linkage in the secondary coil due to the primary coil. Then, using Faraday's law of electromagnetic induction, we can find out the required value of the peak voltage.

Complete step-by-step solution:
We know that inside the coils of a transformer, sinusoidal current flows. Let us consider the current as a function of time by the following expression.
$I = {I_0}\sin \omega t$ ………………...(1)
Now, we know that the flux linked with the secondary coil due to the current in the primary coil is given by
$\varphi = MI$
Substituting (1) in the above expression, we get
$\varphi = M{I_0}\sin \omega t$..................(2)
Now, we know that the magnitude of the voltage induced in a coil is given by
$e = \dfrac{{d\varphi }}{{dt}}$......................(3)
Therefore, differentiating both sides of (2) with respect to the time $t$, we get
$\dfrac{{d\varphi }}{{dt}} = M{I_0}\omega \cos \omega t$......................(4)
Putting (4) in (3) we get
$e = M{I_0}\omega \cos \omega t$
So the peak value of the voltage in the secondary is given by
${e_0} = M{I_0}\omega $.................(5)
According to the question, the peak value of the current is equal to $5{\text{A}}$. So we have
${I_0} = 5{\text{A}}$.........................(6)
Also, the frequency of the current is given to be equal to $50Hz$. So we have
$f = 50Hz$
We know that the angular frequency is related to the frequency by the relation
$\omega = 2\pi f$
$ \Rightarrow \omega = 2\pi \times 50 = 100\pi $.......................(7)
Also, the value of the mutual induction between the primary and the secondary coils of the given transformer is equal to $2H$. So we have
$M = 2H$............................(8)
Substituting (6), (7), and (8) in (5) we ge
${e_0} = 2 \times 5 \times 100\pi {\text{V}}$
$ \Rightarrow {e_0} = 1000\pi {\text{V}}$
Thus, the value of the peak voltage in the secondary coil is equal to $1000\pi $ volts.

Hence, the correct answer is option C.

Note: It is better to remember the expression of the emf induced in the secondary coil of a transformer than to derive it like in the solution above. We have derived the expression in the above solution just for the understanding purpose.