A box weighs 196 ; N on a spring balance at t | vedclass

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(D) The weight of the box at the North Pole is $W_p = mg = 196 \; N$. Given $g = 10 \; m/s^2$,the mass $m = 196 / 10 = 19.6 \; kg$.At the equator,the

Vedclass · Accepted Answer

$195.32$

Vedclass · Answer

(D) The weight of the box at the North Pole is $W_p = mg = 196 \; N$. Given $g = 10 \; m/s^2$,the mass $m = 196 / 10 = 19.6 \; kg$.At the equator,the effective acceleration due to gravity is $g' = g - \omega^2 R$,where $\omega$ is the angular velocity of the Earth.The weight at the equator is $W_e = m(g - \omega^2 R) = mg - m\omega^2 R$.The angular velocity $\omega = \frac{2\pi}{T}$,where $T = 24 	imes 3600 \; s$.Substituting the values: $W_e = 196 - 19.6 	imes \left( \frac{2\pi}{24 	imes 3600} ight)^2 	imes 6400 	imes 10^3$.Calculating the term $m\omega^2 R = 19.6 	imes (7.27 	imes 10^{-5})^2 	imes 6.4 	imes 10^6 \approx 19.6 	imes 0.0337 \approx 0.66 \; N$.Therefore,$W_e = 196 - 0.66 = 195.34 \; N$.The closest option is $195.32 \; N$.

$A$ box weighs $196 \; N$ on a spring balance at the North Pole. Its weight recorded on the same balance if it is shifted to the equator is close to ....... $N$ (Take $g = 10 \; m/s^2$ at the North Pole and the radius of the Earth $R = 6400 \; km$).

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