Question.
The capacity of a closed cylindrical vessel of height $1 \mathrm{~m}$ is $15.4$ litres. How many square metres of metal sheet would be needed to make it?
$\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
$\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Let the radius of the circular end be r.
Height (h) of cylindrical vessel = 1 m
Volume of cylindrical vessel $=15.4$ litres $=0.0154 \mathrm{~m}^{3}$
$\pi r^{2} h=0.0154 \mathrm{~m}^{3}$
$\left(\frac{22}{7} \times r^{2} \times 1\right) \mathrm{m}=0.0154 \mathrm{~m}^{3}$
$\Rightarrow r=0.07 \mathrm{~m}$
Total surface area of vessel $=2 \pi r(r+h)$
$=\left[2 \times \frac{22}{7} \times 0.07(0.07+1)\right] \mathrm{m}^{2}$
$=0.44 \times 1.07 \mathrm{~m}^{2}$
$=0.4708 \mathrm{~m}^{2}$
Therefore, $0.4708 \mathrm{~m}^{2}$ of the metal sheet would be required to make the cylindrical vessel.
Let the radius of the circular end be r.
Height (h) of cylindrical vessel = 1 m
Volume of cylindrical vessel $=15.4$ litres $=0.0154 \mathrm{~m}^{3}$
$\pi r^{2} h=0.0154 \mathrm{~m}^{3}$
$\left(\frac{22}{7} \times r^{2} \times 1\right) \mathrm{m}=0.0154 \mathrm{~m}^{3}$
$\Rightarrow r=0.07 \mathrm{~m}$
Total surface area of vessel $=2 \pi r(r+h)$
$=\left[2 \times \frac{22}{7} \times 0.07(0.07+1)\right] \mathrm{m}^{2}$
$=0.44 \times 1.07 \mathrm{~m}^{2}$
$=0.4708 \mathrm{~m}^{2}$
Therefore, $0.4708 \mathrm{~m}^{2}$ of the metal sheet would be required to make the cylindrical vessel.
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