Question

During motion of a man from equator to pole of earth, its weight will (neglect the effect of change in the radius of earth)

• A. Increase by 0.34%
• B. Decrease by 0.34%
• C. Increase by 0.52%
• D. Decrease by 0.52%

Answer 1

Answer: (A)

$w_{e q}=m g-m \omega^{2} R_{e}$
$w_{p}=m g$
$\frac{w_{p}-w_{e q}}{w_{\theta q}}=\frac{m \omega^{2} R}{m g-m \omega^{2} R}$
$=\frac{\omega^{2} R}{g-\omega^{2} R} $
$\left[\omega^{2} R=0.0337 \,m / s ^{2}\right] $
$ \Rightarrow \frac{\Delta w}{w_{e q}}=\frac{0.0337}{9.81-0.0337}=0.3447 \times 10^{-2}$
$\Rightarrow \frac{\Delta w}{w_{eq}} \times 100 = 0.3447$
$\Rightarrow $ Increases by $0.34 \%$