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
(A) 99.66 N
(B) 110 N
(C) 97.66 N
(D) 106N

Answer 1

Hint: The weight of a body is 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 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: The formulae used in the solution are given here.
$w = mg$ where $w$ is the weight, $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Centrifugal Force ${F_c} = m{\omega ^2}R$ where $m$ is the mass of the body, $\omega $ is the angular velocity and $R$ is the radius of the Earth.
$\omega = 2\pi f$ where $f$ is 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 $w = mg$ where $w$ is the weight, $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Here the reading of the spring balance corresponds to $100N$ i.e. $w = mg = 100N$.
From this we can find the mass of the object as, $m = \dfrac{w}{g} = \dfrac{{100}}{{10}} = 10kg$.
One complete rotation of the Earth around its own axis takes 24 hours.
For a body placed on the equator, it is acted upon by a centrifugal force directed outwards from the axis of rotation due to the rotation of earth. This force is given by the centrifugal force, ${F_c} = m{\omega ^2}R$ where $m$ is the mass of the body, $\omega $ is the angular velocity and $R$ is the radius of the Earth.
Angular velocity is given by $\omega = 2\pi f = \dfrac{{2\pi }}{T}$ where $f$ is the frequency of the rotating body and $T$ is the time period.
Thus, $\omega = \dfrac{{2\pi }}{{86400}}$, since ${\text{1day = 24hrs = 86400s}}$.
Given that, the radius of the earth, $R = 6.4 \times {10^3}m$.
Thus, ${F_c} = m{\omega ^2}R = 10 \times {\left( {\dfrac{{2\pi }}{{86400}}} \right)^2} \times 6.4 \times {10^6} = 0.33N$
Since, the gravitational and centrifugal forces act in different directions, the net force is given by,
${F_{net}} = w - {F_C}$
Assigning the values in the reaction above, ${{\text{F}}_{{\text{net}}}}{\text{ = 100 - 0}}{\text{.33 = 99}}{\text{.66N}}$.

Thus, the answer is Option A.

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 will feel weightlessness, at the centre of earth.