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**Hint:**The time period of a simple pendulum T, depends on the length of the pendulum l and the acceleration due to gravity g. Time period T is directly proportional to the square root of the length of the pendulum l. By equating the two time periods we get the value of n.

**Complete answer:**

According to the question, we have two simple pendulums A and B of length ${{l}_{1}}=5$m and ${{l}_{2}}=20$m respectively, which are given small linear displacement in one direction at the same time, as shown in the diagram below.

Now, let us calculate the time period of pendulum A,

${{T}_{1}}=2\pi \sqrt{\dfrac{{{l}_{1}}}{g}}=2\pi \sqrt{\dfrac{5}{10}}=\sqrt{2}\pi $ ….(i) [ taking acceleration due to gravity $g=10$m/${{s}^{2}}$]

Then, time period of pendulum B is given by,

${{T}_{2}}=2\pi \sqrt{\dfrac{{{l}_{2}}}{g}}=2\pi \sqrt{\dfrac{20}{10}}=2\sqrt{2}\pi $ ….(ii) [ taking acceleration due to gravity $g=10$m/${{s}^{2}}$]

Now, to calculate the number of oscillations n of the pendulum A after which the two pendulums A and B will be in the phase, we compare the time period of pendulum A and time period of pendulum B calculated in equations (i) and (ii), we get,

${{T}_{2}}=2{{T}_{1}}$ ….(iii)

Therefore, we see from the above equation that the time period of pendulum A is twice the time period of pendulum B, i.e. after 2 oscillations of pendulum A, the two pendulums A and B will be in the phase.

**Thus, the correct answer is option (C).**

**Note:**

The formula of the time period of a simple pendulum should be remembered to solve such questions. The standard value of acceleration due to gravity g is $9.8$ m/${{s}^{2}}$, however we can take $g=10$m/${{s}^{2}}$to make the calculations simple and less time taking.

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