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(C) Internal radius,$r = 10 \, cm$.External radius,$R = 12 \, cm$.Volume of metal in the shell $= \frac{4}{3} \pi (R^3 - r^3)$.Volume $= \frac{4}{3} \

Vedclass · Accepted Answer

$14942.28$

Vedclass · Answer

(C) Internal radius,$r = 10 \, cm$.External radius,$R = 12 \, cm$.Volume of metal in the shell $= \frac{4}{3} \pi (R^3 - r^3)$.Volume $= \frac{4}{3} 	imes 3.1416 	imes (12^3 - 10^3) = \frac{4}{3} 	imes 3.1416 	imes (1728 - 1000) = \frac{4}{3} 	imes 3.1416 	imes 728 \, cm^3$.Weight of the shell $= 	ext{Volume} 	imes 	ext{Density}$.Weight $= (\frac{4}{3} 	imes 3.1416 	imes 728) 	imes 4.9 = 3049.4432 	imes 4.9 = 14942.27168 \, gm \approx 14942.28 \, gm$.

$A$ hollow spherical shell is made of a metal of density $4.9 \, g/cm^3$. If its internal and external radii are $10 \, cm$ and $12 \, cm$ respectively,find the weight of the shell. (Take $\pi = 3.1416$) (in $gm$)

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