During motion of a man from equator to pole o | vedclass

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Question

(a)

$w_{e q}=m g-m \omega^2 R_e$

$w_ho=m g$

$\frac{w_p-w_{e q}}{w_{	ext {eq }}}=\frac{m \omega^2 R}{m g-m \omega^2 R}$

$=\frac{\omega^2

Vedclass · Accepted Answer

Increase by $0.34$

Vedclass · Answer

(a)

$w_{e q}=m g-m \omega^2 R_e$

$w_ho=m g$

$\frac{w_p-w_{e q}}{w_{	ext {eq }}}=\frac{m \omega^2 R}{m g-m \omega^2 R}$

$=\frac{\omega^2 R}{g-\omega^2 R} \quad\left[\omega^2 R=0.0337 \,m / s ^2ight]$

$\Rightarrow \frac{\Delta w}{w_{e q}}=\frac{0.0337}{9.81-0.0337}=0.3447 	imes 10^{-2}$

$\Rightarrow \frac{\Delta w}{w_{e q}} 	imes 100=0.3447$

$\Rightarrow 	ext { Increases by } 0.34 \%$

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

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