Question

A body weighs $49\text{N}$ on a spring balance at the North Pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

$[$Use, $g=\frac{GM}{R^2}=9.8{\text{m/s}}^2$ and radius of earth, $R=6400\text{km}]$

• A. $49\text{N}$
• B. $49.83\text{N}$
• C. $49.17\text{N}$
• D. $48.83\text{N}$

Answer 1

The correct option is D $48.83\text{N}$

Let the mass of the body be $M$.

So, weight at the North Pole, $Mg=49\text{N}$

Now, at the equator, gravitational acceleration is less than that at the North Pole due to rotation of the earth, and it is given as:
$g^{\prime }=g-{\omega }^{2}R$

Where, $\omega$ is angular speed of rotation of earth, $R$ is radius of the earth and $g$ is gravitational acceleration at the pole.

Now, Weight at the equator, ${Mg}^{\prime }=M(g-{\omega }^{2}R)$

So, weight at the equator will be less than that at the North Pole.

From the given option, only fourth option is having value lesser than $49\text{N}$.