Question

A body weighs $1$ kg by a spring balance at the north pole. What will it weight at the equator?

• A. $0.977kg$
• B. $0.967kg$
• C. $0.987kg$
• D. $0.997kg$

Answer 1

The correct option is D $0.997kg$

At north pole, $\omega =0$, therefore, weight is mg=1

At the equator, $F=mg-mR{\omega }^2$

$\omega =\frac{2\pi }{3600\times 24}=7.27\times {10}^{-5}rad/s$

Therefore, force at the equator is $mg(1-\frac{R{\omega }^2}{g})$

$\Rightarrow F=1(1-\frac{6400\times 1000\times {(7.27\times {10}^{-5})}^2}{10})=0.997kg$