JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Gravitation

At the pole the balance reads the full gravitational pull, so the mass is $m=\frac{49}{9.8}=5$ kg. At the equator the body is carried round a circle of radius $R$ by the earth's rotation, and part of the gravitational pull goes into the centripetal acceleration, leaving an effective

With $\omega =\frac{2\pi }{86400}=7.27\times 10^{-5}\mathrm{rad}{\mathrm{s}}^{-1}$ and $R=6.4\times 10^6$ m, ${\omega }^{2}R=0.034\mathrm{m}{\mathrm{s}}^{-2}$, so $g^{\prime }=9.8-0.034=9.766\mathrm{m}{\mathrm{s}}^{-2}$. The weighing machine at the equator therefore records $5\times 9.766=48.83$ N.