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**Hint:**To solve this question, we need to know about the relationship between the length of a pendulum and the number of oscillations that it could complete in a certain time period and finally calculate the number of oscillations of the shorter pendulum after which both the pendulums will be in phase again.

**Complete step-by-step solution:**

In the question we are given two pendulums that have different lengths and are given small linear displacement in the same direction at the same time. We know the relationship between the length of a pendulum and its time period. The time period of a pendulum is directly proportional to the square root of its length. Hence, we can write the relation as follows:

$T\propto \sqrt{l}$

On removing the proportionality sign, we get:

$T=2\pi \sqrt{\dfrac{l}{g}}$

From this relation we can conclude that the time taken by a particular pendulum to make a single oscillation would depend only on its length as all the other values are constants. We can also say that the number of oscillations that a particular pendulum makes in a given time period would depend only on its length. Now, let us calculate the time period of the simple pendulums of length $1\text{ m}$ and $4\text{ m}$:

${{T}_{1}}=2\pi \sqrt{\dfrac{1}{g}}$

And

$\begin{align}

& {{T}_{2}}=2\pi \sqrt{\dfrac{4}{g}} \\

& \Rightarrow {{T}_{2}}=4\pi \sqrt{\dfrac{1}{g}} \\

\end{align}$

Thus, we can say that:

${{T}_{2}}=2{{T}_{1}}$

Thus, we can conclude that the time period of the second pendulum that has a length of $4\text{ m}$ takes twice the time taken by the first pendulum that has a length of $1\text{ m}$. Hence, the shorter pendulum will make two oscillations by the time it coincides with the longer pendulum. Hence the correct option is $C$.

**Note:**As we saw above, the lengths of the strings of pendulums is the only factor that can vary their time period of oscillation and nothing else because the time period of a pendulum is directly proportional to the square root of its length. Thus, any other factors like displacement from mean position do noy affect the time period of a pendulum.

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