The period of the function $f(x) = \sin \left( {\sin \dfrac{x}{5}} \right)$ is
A) $2\pi $
B) $\dfrac{{2\pi }}{5}$
C) $10\pi $
D) $5\pi $

Hint: We should first know what a periodic function is. Periodic function is nothing but a function which repeats its value at a regular interval of time. We have an equation for the periodicity of function given by,
 $f(x + T) = f(x)$
To solve the above question we should know the concept that if
$f(x)$ is a periodic function with period T and
$g(x)$ is any function such that range if $f$ is a proper subset of the domain $g$, then we can say that $g(f(x))$ is aperiodic with period $T$ .

Complete step by step answer:
In the given question we have
$f(x) = \sin \left( {\sin \dfrac{x}{5}} \right)$
If we observe the terms in the given function they are in the form of
$f(g(x))$ which is a composite function.
We know that the period of a function of type $f(g(x))$ is the same as the period of
$g(x)$ .
By comparing from the question, we have
$g(x) = \sin \left( {\dfrac{x}{5}} \right)$
Now we know that the period of sine function is $2\pi $
But in the above function we have a denominator i.e. divided by $5$ .
So to make the function as $2\pi $, we have to multiply it with $5$
So we have
$\sin \left( {\dfrac{x}{5}} \right) = 2\pi \times 5 = 10\pi $
We can write this as
 $f(x) = \sin \left( {\sin \dfrac{x}{5}} \right)$
Now from the above concept we can say that the period of $g(x) = 10\pi $ , and we know that the period of a function of type $f(g(x))$ is the same as the period of $g(x)$ .
So we can say that
$ \therefore f(x) = \sin \left( {\sin \dfrac{x}{5}} \right) = 10\pi $. Hence, optio (C) is correct.

Note:
We should note that the period of the function of cosine i.e. $\cos x$ is also
 $2\pi $. We should know that we can always calculate the period using the formula derived from the basic sine and cosine equation.
The period for function $y = A\sin (Ba - c)$ and $Y = A\sin (Ba - c)$ is equal to
$2\pi B$ radians.