The acceleration due to gravity increases by 0.5% 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

The correct option is 2.005 s.

Step-by-step Explanation:

A pendulum that "beats seconds" is a seconds pendulum, which has a time period of 2 seconds. Thus, its time period at the poles ($T_p$) is:
$T_p=2\text{s}$

Let the acceleration due to gravity at the equator be $g_e$ and at the poles be $g_p$. Since the acceleration due to gravity increases by 0.5% from the equator to the poles, we can write:
$g_p=g_e+0.5\%\text{of}{g}_e=g_e\left(1+0.005\right)=1.005g_e$

The time period ($T$) of a simple pendulum of length $l$ is given by the formula:
$T=2\pi \sqrt{\frac{l}{g}}$

From this formula, we can see that the time period is inversely proportional to the square root of the acceleration due to gravity (assuming the length $l$ of the pendulum remains constant):
$T\propto \frac{1}{\sqrt{g}}$

Therefore, the ratio of the time period at the equator ($T_e$) to that at the poles ($T_p$) is:
$\frac{T_e}{T_p}=\sqrt{\frac{g_p}{g_e}}$

Substitute the relationship between $g_p$ and $g_e$ into the equation:
$\frac{T_e}{T_p}=\sqrt{\frac{1.005g_e}{g_e}}=\sqrt{1.005}={\left(1+0.005\right)}^{\frac{1}{2}}$

Using the binomial approximation ${\left(1+x\right)}^n\approx 1+nx$ (since $x=0.005$ is much smaller than 1):
$\frac{T_e}{T_p}\approx 1+\frac{1}{2}\left(0.005\right)=1+0.0025=1.0025$

Now, solve for the time period at the equator $T_e$:
$T_e\approx 1.0025\times T_p$

Substitute the value $T_p=2\text{s}$:
$T_e\approx 1.0025\times 2\text{s}=2.005\text{s}$

Thus, the time period of the pendulum at the equator is 2.005 s.