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**Hint:**The product of mass of the body and the acceleration due to gravity is called the weight of the body.Acceleration due to gravity varies with the angular velocity of rotation of the earth as $g' = g - {\omega ^2}R\cos \theta $. But at the poles the acceleration due to gravity does not change with $\omega $ as $\theta = 0^\circ $. Using this information, we can find out the correct answer.

**Complete answer:**

Let us assume a test mass m. the force of gravity acting on it is given as mg. The centrifugal force on this test charge is $m{\omega ^2}r$.As both these forces are co-initial as well as coplanar forces, we can add them vectors using parallelogram law of vector addition to get the magnitude of the apparent value of the gravitational force as,

${\left( {mg'} \right)^2} = {\left( {mg} \right)^2} + {\left( {m{\omega ^2}r} \right)^2} + 2(mg)(m{\omega ^2}r)\cos (\pi - \theta )$

Where r is the radius of the circular path and ‘R’ is the radius of the earth, then $r = R\cos \theta $.

Substituting the value of r we get,

$g' = g - R{\omega ^2}\cos (2\theta )$

Where g′ is the apparent value of acceleration due to gravity at the latitude due to the rotation of the earth and g is the true value of gravity at the latitude without considering the rotation of the earth.

At poles, $\theta = 90^\circ \Rightarrow g' = g$.

At the equator,$\theta = 0^\circ \Rightarrow g' = g - {\omega ^2}R$ .

**Hence, option D is the correct answer.**

**Note:**The gravitational acceleration or acceleration due to gravity is denoted by the symbol ‘g’ and has a standard value of 9.8 m/s but it varies with change in gravitational environments. The value of g is more at equator and is constant at poles and it decreases with the increase in the angular velocity of rotation of the earth. Acceleration due to gravity decreases with increase in angular velocity, $\omega $ as the acceleration due to gravity decreases, the weight of the body being directly proportional to it also decreases.

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