Question.
A hemispherical tank is made up of an iron sheet $1 \mathrm{~cm}$ thick. If the inner radius is $1 \mathrm{~m}$, then find the volume of the iron used to make the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
A hemispherical tank is made up of an iron sheet $1 \mathrm{~cm}$ thick. If the inner radius is $1 \mathrm{~m}$, then find the volume of the iron used to make the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Inner radius $\left(r_{1}\right)$ of hemispherical tank $=1 \mathrm{~m}$
Thickness of hemispherical tank $=1 \mathrm{~cm}=0.01 \mathrm{~m}$
Outer radius $\left(r_{2}\right)$ of hemispherical tank $=(1+0.01) \mathrm{m}=1.01 \mathrm{~m}$
Volume of iron used to make such a tank $=\frac{2}{3}\left(r_{2}^{3}-r_{1}^{3}\right)$
$=\left[\frac{2}{3} \times \frac{22}{7} \times\left\{(1.01)^{3}-(1)^{3}\right\}\right] \mathrm{m}^{3}$
$=\left[\frac{44}{21} \times(1.030301-1)\right] \mathrm{m}^{3}$
$=0.06348 \mathrm{~m}^{3} \quad$ (approximately)
Inner radius $\left(r_{1}\right)$ of hemispherical tank $=1 \mathrm{~m}$
Thickness of hemispherical tank $=1 \mathrm{~cm}=0.01 \mathrm{~m}$
Outer radius $\left(r_{2}\right)$ of hemispherical tank $=(1+0.01) \mathrm{m}=1.01 \mathrm{~m}$
Volume of iron used to make such a tank $=\frac{2}{3}\left(r_{2}^{3}-r_{1}^{3}\right)$
$=\left[\frac{2}{3} \times \frac{22}{7} \times\left\{(1.01)^{3}-(1)^{3}\right\}\right] \mathrm{m}^{3}$
$=\left[\frac{44}{21} \times(1.030301-1)\right] \mathrm{m}^{3}$
$=0.06348 \mathrm{~m}^{3} \quad$ (approximately)
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