The acceleration due to gravity increases by $0.5$ percent when we go from the equator to the poles. What will be the time period of the pendulum at the equator which beats seconds at the poles.
(A) $1.950s$
(B) $1.995s$
(C) $2.050s$
(D) $2.005s$

Hint:In order to solve this question, we will first calculate the actual value of acceleration due to gravity at poles in terms of ‘g’, and then using the general formula of the time period of the pendulum, we will solve for the time period of the pendulum at the equator if its time period is a second at poles.

Formula used:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
Where,
g-The acceleration due to gravity at some place
l-The length of the pendulum then

Complete answer:
Let us assume acceleration due to gravity at the equator is g and at the poles, it’s increased by 0.5%, the value of gravity at pole g’ is calculated as
$
  g' = g + \dfrac{{0.5g}}{{100}} \\
  g' = 1.005g \\
 $

Now, time period of the pendulum at equator is given by $T = 2\pi \sqrt {\dfrac{l}{g}} $ and time period at the pole, we have given that it beats seconds which means its time period will be
$T' = 2\pi \sqrt {\dfrac{l}{{g'}}} = 2s$

on putting the value of g’ we get,
$
  2\pi \sqrt {\dfrac{l}{{g'}}} = 2s \\
  2\pi \sqrt {\dfrac{l}{{(1.005)g}}} = 2 \\
  2\pi \sqrt {\dfrac{l}{g}} = 2\sqrt {1.005} \\
 $

Also, we have time period at equator is $T = 2\pi \sqrt {\dfrac{l}{g}} $ so, we get,
$
  T = 2\sqrt {1.005} \\
  T = 1.995s \\
 $

So, the time period of the pendulum at the equator is $1.995s$, option B.

Note: It should be remembered that as long as the length of the pendulum keeps constant the time period of the pendulum varies with places due to the non-constant value of acceleration due to gravity on earth.