Question

# A hemi-spherical tank is made up of an iron sheet $1cm$ thick. If the inner radius is$1m$, then find the volume of the iron used to make the tank. (Assume $\mathrm{\pi }=\frac{22}{7}$)

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Solution

## Inner Radius of the tank, (r) $=1m$The thickness of the hemispherical tank $=1cm=0.01m$Therefore, the outer Radius (R) $=1+0.01=1.01m$Volume of the iron used in the hemi-spherical tank$=\frac{2}{3}\mathrm{\pi }\left({\mathrm{R}}^{3}-{\mathrm{r}}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{3}×\frac{22}{7}×\left(1.{01}^{3}-{1}^{3}\right)\phantom{\rule{0ex}{0ex}}=0.06348{\mathrm{m}}^{3}$Hence, the volume of the iron used in the hemispherical tank is $0.06348{m}^{3}$

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