Question

The acceleration due to gravity increases by 0.5% when we go from the equator to the poles

Answer 1

Answer:

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.

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Answer 2

Given that,

The acceleration due to gravity increases by 0.5% when we go from the equator to the poles

The gravity at poles is

$g'=g+0.5g\%$

$g'=g(1.005)$

Suppose, what will be the time period of the pendulum at the equator which beats seconds at the poles

We need to calculate the time period of the pendulum at the equator

Using formula of time period

The time period at equator is

$T=2\pi\sqrt{\dfrac{l}{g}}$....(I)

The time period at poles is

$T'=2\pi\sqrt{\dfrac{l}{g'}}$

Put the value of g'

$T'=2\pi\sqrt{\dfrac{l}{g(1.005)}}$....(II)

$T'=\dfrac{T}{\sqrt{1.005}}$

$T=T'\times 1.00249$

$T=1.00249T'$

Hence, The time period of the pendulum at the equator is 1.00249 times of time period at the pole.