A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is $ \frac{14}{2} \mathrm{~m} $ and the diameter of hemisphere is $ 3.5 \mathrm{~m} $. Calculate the volume and the internal surface area of the solid

Given:

A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base.

The depth of the cylinder is \( \frac{14}{2} \mathrm{~m} \) and the diameter of hemisphere is \( 3.5 \mathrm{~m} \).

To do:

We have to find the volume and the internal surface area of the solid.

Solution:

Diameter of the hollow cylinder $= 3.5\ m$

This implies,

Radius of the cylinder $r = \frac{3.5}{2}$

$ =\frac{7}{4}\ m$

Height of the cylindrical part $h = \frac{14}{3}\ m$

Volume of the solid $=$ Volume of the cylindrical part $+$ Volume of the hemispherical part

$=\pi r^{2} h+\frac{2}{3} \pi r^{3}$

$=\pi r^{2}(h+\frac{2}{3} r)$

$=\frac{22}{7} \times(\frac{7}{4})^{2}(\frac{14}{3}+\frac{2}{3} \times \frac{7}{4})$

$=\frac{22 \times 49}{7 \times 16}(\frac{14}{3}+\frac{7}{6})$

$=\frac{77}{8}(\frac{28+7}{6})$

$=\frac{77}{8} \times \frac{35}{6}$

$=\frac{2695}{48}$

$=56.1458$

$=56.15 \mathrm{~m}^{3}$

Internal surface area of the solid $=2 \pi r h+2 \pi r^{2}$

$=2 \pi r(h+r)$

$=2 \times \frac{22}{7} \times \frac{7}{4}(\frac{14}{3}+\frac{7}{4})$

$=11(\frac{56+21}{12})$

$=11 \times \frac{77}{12}$

$=\frac{847}{12}$

$=70 \frac{7}{12} \mathrm{~m}^{2}$

The volume and the internal surface area of the solid are $56.15 \mathrm{~m}^{3}$ and $70 \frac{7}{12} \mathrm{~m}^{2}$.

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Updated on: 10-Oct-2022

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