Question

A cylindrical tank of radius 80 cm contains water to a
depth of 2 m. What is the total area of wetted surface?​

Answer 1

Given -

A cylindrical tank of radius 80 cm contains water to a depth of 2 m.

To find -

• Total surface area of wetted surface

Solution -

• Radius of cylindrical tank = 80 cm
• Height or depth of cylindrical tank = 2 m

Focus Zone

• 1 m = 100 cm

→ Radius = 80 cm = 80/100 = 0.8 m

According to the formula of total surface area of cylinder

→ Curved surface of cylinder + 2 × area of circle

→ 2πrh + 2πr²

→ 2πr(h + r)

Focus Zone : Don't include area of two circle. It's because we have to find wetted surface area of cylindrical tank.

→ 2πrh + πr²

→ πr(2h + r)

→ 22/7 × 0.8(2 × 2 + 0.8)

→ 17.6/7(4 + 0.8)

→ 17.6/7 × 4.8

→ 84.48

→ 12.06 m²

•°• Total surface area of wetted surface = 12.06 m²

________________________________

Answer 2

$\; \; \; \; \; \; \; \;{\Large{\bf{\underbrace{Required \; Solution}}}}$

${\large{\bold{\sf{\underline{Given \; that}}}}}$

★ Radius of cylindrical tank = 80 cm or 0.8 metres.

Note - As we already know that 1 m = 100 cm so 1 cm = 1/100 metres.

★ Tank contains water at depth = 2 m.

${\large{\bold{\sf{\underline{To \; find}}}}}$

★ The total area of wetted surface

${\large{\bold{\sf{\underline{Solution}}}}}$

★ The total area of wetted surface = 12.06 cm²

${\large{\bold{\sf{\underline{Using \; concept}}}}}$

★ Formula to find TSA (total surface area) of cylinder.

${\large{\bold{\sf{\underline{Using \; formula}}}}}$

◉ TSA (cylinder) = CSA + 2 × Area of circle

So, let's see

◉ CSA (cylinder) = 2πrh

◉ Area (circle) = 2πr²

Therefore,

◉ 2πr(h+r)

➝ πr(2h+r)

➝ 22/7(0.8) (2(2) + 0.8)

➝ 22/7 × 0.8 (2 × 2 + 0.8)

➝ 22/7 × 0.8 (4+0.8)

➝ 17.6/7 (4+0.8)

➝ 17.6/7 (4.8)

➝ 17.6/7 × 4.8

➝ 84.48/7

➝ 12.06 cm²

${\large{\bold{\sf{\underline{Additional \; information}}}}}$

Cylinder diagram -

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{r}}\put(9,17.5){\sf{h}}\end{picture}$

Formula -

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \: \dfrac{v}{\pi h}}}}$

Diagram of this question -

Kindly see from attachment