A student says that he had applied a force $F=-k\sqrt{x}$ on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle.
A. As x increases k increases.
B. As x increases k decreases.
C. As x increases k remains constant.
D. The motion cannot be simple harmonic.

Hint: This is a tricky question as the value of force is an equation and the relation of the variables that is asked in the question is mentioned in the equation only. So, to get the answer for this type of question, first substitute the value of force that is applied in simple harmonic motion and then equate the given equation with the known equation, which will help to get the relation that will help to answer the question.

Formula used:
$F=-m{{\omega }^{2}}x$

Complete solution:
It is given in the question that the particle is moving in Simple Harmonic Motion, due to which the force applied on the particle is given by:
$F=-m{{\omega }^{2}}x$
And the force applied by student is given as:
$F=-k\sqrt{x}$
So, if the motion is Simple Harmonic Motion, then the value of both the force equation must be equal, which means,
$\begin{align}
  & -m{{\omega }^{2}}x=-k\sqrt{x} \\
 & \Rightarrow m{{\omega }^{2}}x=k\sqrt{x} \\
 & \text{Squaring on both sides, we get -} \\
 & \Rightarrow {{m}^{2}}{{\omega }^{4}}{{x}^{2}}={{k}^{2}}x \\
 & \therefore {{k}^{2}}={{m}^{2}}{{\omega }^{4}}x \\
\end{align}$
From the above relation, it is clear that as x increases k increases as well.

Therefore, the correct answer is Option (A).

Note:
To solve this type of questions, it's better to use an existing equation and then equate the given and the known equations to get the relation. There is no better way other than this to get the solution of this type of question, as there is not much information given in the question that can help to get the relation.