Question

The reading of a spring balance corresponds to 100N while situated at the North Pole and a body is kept on it. The weight record on the same scale if it is shifted to the equator, is
(Take,$g = 10m/{s^2}$and radius of the earth, $R = 6.4 \times {10^6}m$)
(A) $99.66N$
(B) $110N$
(C) $97.66N$
(D) $106N$

Answer 1

We know that the weight of a body can be defined as the gravitational force exerted by the earth on a body. It depends on the mass of the body as well as the acceleration due to gravity. The mass of the body is always constant and we consider the acceleration due to gravity also to be a constant. Now the weight of the body might change when its position is shifted due to the presence of some other force acting on the body.
Formula used:
$w = mg$ (Where $w$stands for the weight of the body, $m$ stands for the mass of the body, and $g$stands for the acceleration due to gravity)
The centrifugal force,
${F_c} = m{\omega ^2}R$(Where ${F_c}$ stands for the centrifugal force, $m$stands for the mass of the body, $\omega$ stands for the angular velocity and $R$stands for the radius of the earth)
$\omega = 2\pi f$(Where $\omega$ stands for the angular velocity, $2\pi$ is a constant and $f$stands for the frequency of the rotating body)

Complete step by step solution:
The weight of an object is the force experienced by an object due to the gravitational pull of the earth. This force is directed towards the center of the earth.
We can write the weight of an object as $w = mg$
Here the reading of the spring balance corresponds to $100N$
i.e. $mg = 100N$
From this we can find the mass of the object as,
$m = \dfrac{{100N}}{g} = \dfrac{{100N}}{{10}} = 10kg$ ($g$ is given as $10m/{s^2}$)
When the body is placed on the equator, it will experience a centrifugal force due to the rotation of earth about its own axis. This force will be directed outwards from the axis of rotation.
This force can be written as,
${F_c} = m{\omega ^2}R$
We know that the mass of the body is $10kg$
Now we need to obtain the angular velocity of the earth.
We know that the time taken for one complete rotation of the earth is given by,
$T = 24hrs$
Converting into seconds,
$T = 24 \times 60 \times 60 = 86400s$
We know that angular velocity can be written as $\omega = \dfrac{{2\pi }}{T}$
The radius of the earth is given by, $R = 6.4 \times {10^6}m$
By putting these values in the expression for the centrifugal force, we get
${F_c} = 10 \times {\left( {\dfrac{{2\pi }}{{86400}}} \right)^2} \times 6.4 \times {10^6} = 0.33N$
We know that the gravitational force $mg$and the centrifugal force ${F_c}$acts in opposite directions. Therefore we can write the net force on the body as,
$F = mg - {F_c}$
$F = 100N - 0.33N = 99.66N$

The answer is Option (A): $99.66N$

Note:
The value of acceleration due to gravity varies with respect to the position. The value of acceleration due to gravity can change with altitude, latitude depth and rotation of the earth. As the altitude increases the value of acceleration due to gravity decreases. Acceleration due to gravity decreases as the depth increases. $g$ is zero at the centre of the earth. Hence we feel weightlessness, at the centre of earth.