Question.
Find
(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) How much steel was actually used, if $\frac{1}{12}$ of the steel actually used was wasted in making the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) How much steel was actually used, if $\frac{1}{12}$ of the steel actually used was wasted in making the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Height (h) of cylindrical tank = 4.5 m
Radius $(r)$ of the circular end of cylindrical tank $=\left(\frac{4.2}{2}\right) \mathrm{m}=2.1 \mathrm{~m}$
(i) Lateral or curved surface area of tank $=2 \pi r h$
$=\left(2 \times
\frac{22}{7} \times 2.1 \times 4.5\right) \mathrm{m}^{2}$
$=(44 \times 0.3 \times 4.5) \mathrm{m}^{2}$
$=59.4 \mathrm{~m}^{2}$
Therefore, CSA of tank is $59.4 \mathrm{~m}^{2}$.
(ii) Total surface area of tank $=2 \pi r(r+h)$
$=\left[2 \times \frac{22}{7} \times 2.1 \times(2.1+4.5)\right] \mathrm{m}^{2}$
$=(44 \times 0.3 \times 6.6) \mathrm{m}^{2}$
$=87.12 \mathrm{~m}^{2}$
Let $\mathrm{A} \mathrm{m}^{2}$ steel sheet be actually used in making the tank.
$\therefore \mathrm{A}\left(1-\frac{1}{12}\right)=87.12 \mathrm{~m}^{2}$
$\Rightarrow \mathrm{A}=\left(\frac{12}{11} \times 87.12\right) \mathrm{m}^{2}$
$\Rightarrow \mathrm{A}=95.04 \mathrm{~m}^{2}$
Therefore, $95.04 \mathrm{~m}^{2}$ steel was used in actual while making such a tank.
Height (h) of cylindrical tank = 4.5 m
Radius $(r)$ of the circular end of cylindrical tank $=\left(\frac{4.2}{2}\right) \mathrm{m}=2.1 \mathrm{~m}$
(i) Lateral or curved surface area of tank $=2 \pi r h$
$=\left(2 \times
\frac{22}{7} \times 2.1 \times 4.5\right) \mathrm{m}^{2}$
$=(44 \times 0.3 \times 4.5) \mathrm{m}^{2}$
$=59.4 \mathrm{~m}^{2}$
Therefore, CSA of tank is $59.4 \mathrm{~m}^{2}$.
(ii) Total surface area of tank $=2 \pi r(r+h)$
$=\left[2 \times \frac{22}{7} \times 2.1 \times(2.1+4.5)\right] \mathrm{m}^{2}$
$=(44 \times 0.3 \times 6.6) \mathrm{m}^{2}$
$=87.12 \mathrm{~m}^{2}$
Let $\mathrm{A} \mathrm{m}^{2}$ steel sheet be actually used in making the tank.
$\therefore \mathrm{A}\left(1-\frac{1}{12}\right)=87.12 \mathrm{~m}^{2}$
$\Rightarrow \mathrm{A}=\left(\frac{12}{11} \times 87.12\right) \mathrm{m}^{2}$
$\Rightarrow \mathrm{A}=95.04 \mathrm{~m}^{2}$
Therefore, $95.04 \mathrm{~m}^{2}$ steel was used in actual while making such a tank.