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Sagot :
Solution :
Volume of ball = Volume of outer ball - Volume of inner ball
[tex]V = \dfrac{4\pi R^3}{3} - \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\pi}{3}( R^3 - r^3 )\\\\V = \dfrac{4\pi}{3}( 200^3 - 100^3 )\\\\V = \dfrac{4\pi}{3}(2^3 - 1^3) \times 10^6\\\\V = 2.93 \times 10^7 \ m^3[/tex]
Now, density of iron is, d = 7300 kg/m³.
So, mass of iron ball is :
[tex]m = d\times V\\\\m = 7300\times 2.93 \times 10^7 \ kg\\\\m = 2.14 \times 10^{11}\ kg[/tex]
Hence, this is the required solution.
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