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 $\left.\pi=\frac{22}{7}\right]$


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.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now