Question

A person is standing at a latitude which makes an angle of ${60}^{\circ }$ with the centre of earth. What is the % change in his actual weight here. Take reference of weight near earth's pole as actual weight. Take $g=10{\text{m/s}}^2$.

• A. $8$%
• B. $0.08$%
• C. $0.008$%
• D. $0.0008$%

Answer 1

The correct option is B $0.08$%

Centrifugal force acting on the person at latitude $\theta$ is $m{\omega }^{2}R\cos \theta$
where $\omega =$ angular velocity of earth's rotation
$R=$ radius of earth

Therefore at latitude $\theta$:
$N+m{\omega }^{2}R{\cos }^2\theta =mg$
$N=mg-m{\omega }^{2}R{\cos }^2\theta$

At the poles , $N=mg$

% change in weight$=\frac{mg-N}{mg}\times 100$
$=\frac{m{\omega }^{2}R{\cos }^2\theta }{mg}\times 100$
$=\frac{R{\omega }^2{\cos }^2\theta }{g}\times 100$
$=\frac{4{\pi }^2\times 6.4\times {10}^6\times 1}{{24}^2\times {3600}^2\times 10\times 4}\times 100$

$\therefore$% change in weight $=0.08$%