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Q.What is the decrease in weight of a body of mass 500 Kg when it is taken into a mine of depth 1000 Km? (Radius of earth R = 6400 km, g=9.8 m/s2g = 9.8\ \text{m/s}^2)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 3mImportance★★★★★
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Below the Earth's surface, gg decreases linearly with depth: gd=g(1−dR)g_d = g\left(1-\dfrac{d}{R}\right), so the weight lost is mg⋅d/Rmg\cdot d/R.

The acceleration due to gravity at depth dd below the Earth's surface (assuming uniform density) is

gd=g(1−dR),g_d = g\left(1 - \frac{d}{R}\right),

where RR is the Earth's radius. So the weight of the body at depth dd is Wd=mgd=mg(1−dR)W_d = mg_d = mg\left(1-\dfrac{d}{R}\right), and the decrease in weight compared to the surface weight mgmg is

ΔW=mg−mgd=mg⋅dR.\Delta W = mg - mg_d = mg\cdot\frac{d}{R}. …

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