The time period of a second’s pendulum on the surface of moon will be
$
  A.\dfrac{2}{{\sqrt 6 }}s \\
  B.6\sqrt 2 s \\
  C.2\sqrt 6 s \\
  D.2s \\
 $

Hint: In order to solve this question, firstly we will write out the equation for the time period on the earth, then again for the time period on the moon earth by using the formula of time period at the surface of moon i.e. $T = 2\pi \sqrt {\dfrac{L}{g}} $.

Complete step-by-step solution -

We know that a seconds pendulum is a pendulum which has a time period of 2 seconds; one second for a swing in one direction and one second for the return swing with a frequency of $\dfrac{1}{2}Hz$.
On earth, the time period will be-$2 = 2\pi \sqrt {\dfrac{l}{g}} .............\left( 1 \right)$
On the surface of the moon, acceleration due to gravity is $\dfrac{g}{6}$.
( value of g at moon is$\dfrac{1}{6}$ of value of g at earth’s surface)
On the surface of the moon, the time period of a second’s pendulum will be$2\pi \sqrt {\dfrac{l}{{\dfrac{g}{6}}}} .............\left( 2 \right)$
Dividing the above equations 1 and 2,
We get-
$\dfrac{{2}}{T_{moon}}$= $\dfrac {{2\pi \sqrt {\dfrac{{l}}{g}}}} {2\pi \sqrt {\dfrac{{l}}{\dfrac{{g}}{6} }}}$
${T_{moon}} = 2\sqrt 6 s$
Therefore, the time period of a second’s pendulum on the surface of the moon will be $2\sqrt 6 s$.
Hence, option C is correct.

Note- While solving this question, we must know that a pendulum with any other value of period cannot be called a seconds pendulum. Thus a seconds pendulum anywhere has a period of 2 seconds, although it’s length will be different at different locations.