If $f(x) = \sin \sqrt x $, then the period of $f(x)$ is:
A. $\pi $
B. $\dfrac{\pi }{2}$
C. $2\pi $
D. None of these

Hint: As we know that trigonometry is one of the most significant branches in mathematics that finds huge use in distinct fields which predominantly deals with the study of the association between the sides and angles of the right-angle triangle. Hence, it helps to determine the misplaced or unbeknown sides or angles of a right triangle by applying trigonometric formulas.
In order to solve this question, we must first assume that the function is periodic, in which case we obtain $\sin x = 2\pi $, here $\sqrt x \geqslant 0$then the values of $x = 0,\pi ,2\pi ,4\pi ,...$By squaring the two equations, we will arrive at an equation that will result in a contradiction. As a result, we establish that the function is not periodic.

Formula Used:
 $\sin x = 2\pi $

Complete step by step Solution:
In the question, the function is given by:
 $f(x) = \sin \sqrt x $
As we know$\sqrt x \geqslant 0$, which has a period $\sin x = 2\pi $, then we have:
$\sqrt x = 0,\,2\pi ,\,4\pi ,\,....$
Squaring both the sides of the above equation, then:
$x = 0,\,4{\pi ^2},\,16{\pi ^2},\,..... \\$
$\Rightarrow x = 0,40,160,..... \\$
Since the quantity or the value inside the sin function varies only from $0$ to $\infty $
There this does meet the condition which is required for the periodic function which is $ - \infty $ to $\infty $.
Thus, the function $f(x) = \sin \sqrt x $ is not periodic.

Hence, the correct option is (D).

Note: It should be noted that the periodic function's domain is always $( - \infty ,\infty )$. However, in this instance, the domain of the function is $( - \infty ,\infty )$. As a result, we may state categorically that the function is neither periodic nor does it have a period. Keep in mind that the opposite of the statement—that is, that not every function with the domain $( - \infty ,\infty )$ is periodic—is not true.