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Answer :
Given :
Inner radius of hemispherical tank (r) = 1 m
Thickness of hemispherical tank 1 cm = 0.01 m
Outer radius of hemispherical tank (R) = (1 + 0.01) m = 1.01 m
We know that,
Volume of iron is used to make such a tank
= \(\frac{2}{3}{\pi}({R}^3 - {r}^3)\)
= \(\frac{2}{3} × \frac{22}{7} × ({1.01}^3 - {1}^3)\) \({m}^3\)
= \(\frac{44}{21} × (1.030301 - 1)\) \({m}^3\)
= 0.06348 \({m}^3\) (approx.)
Therefore, the volume of the iron used to make the tank is 0.06348 \({m}^3\).