Question

If weight of an object at pole is $196N$ then weight at equator is

Answer 1

$g = 10m{s^{ - 2}}$, radius of earth is $6400Km$, weight at equator and pole difference so take angular velocity time then eventually you will get the apparent weight by putting the formula.
Formulae used:
$\omega = \dfrac{{2\pi }}{T}$
Where $\omega$ is the angular velocity of
$T$ is the time
$2\prod$ is the constant
$Mg' = mg - m{\omega ^2}R$
Where $m$ is the weight given in question
$mg$ is the gravitational force towards the centre of earth
$\omega$ is the angular velocity of earth
R is the radius of earth
$Mg'$ is the apparent weight

Complete step by step solution:
Earth rotates at its axis at an angular point speed of a revolution per 24 hours
$m{\omega ^2}R$ is the centrifugal force which drives the object away from the centre
Gravitational force towards the centre of the earth $mg$ that is the force is vertically downward
at $\phi = 90^\circ$ at equator
So coming to the solution,
Weight of an object at pole is $196N$
Gravitational force of earth is $9.8m{s^{ - 2}}$ to solve the solution we can use approx. value that is $g = 10m{s^{ - 2}}$
$w = mg$
$\Rightarrow 196 = m \times 10$
$\Rightarrow$ $m = 19.6kg$
$19.6Kg$ is the weight at the poles of earth.
While, it revolves around the sun so at the equator we get an
Radius of the earth is $6400Km = 6400 \times {10^3}$
Obvious weight$= mg - m{\omega ^2}R$
$\omega = \dfrac{{2\pi }}{T}$
$T = 24hours = 24 \times 3600\sec$
$\Rightarrow \dfrac{{2 \times 3.14}}{{24 \times 3600}}$
$\Rightarrow \dfrac{{2 \times 3.14}}{{24 \times 3600}}$
Apparent weight$= 19.6 - 19.6{(\dfrac{{2 \times 3.14}}{{24 \times 3600}})^2} \times 6400 \times {10^3}$
$= 195.33N$
$195.33N$ is the weight at the equator.

Additional Information:
The equator is more far from the centre of earth as compared to that of the distance of poles. The effective acceleration of gravity at the poles is $980.665cm{s^{ - 1}}$ while at the equator it is $3.39cm{s^{ - 1}}$ less due to the centrifugal force.

Note:
While using and changing the SI units need to be careful and while using the sign in the equation. Weight of the equator is less than the pole so you can verify your answer.