Question

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

Answer 1

Answer:-

$\red{\bigstar}$ Area of wetted surface $\large\leadsto\boxed{\tt\green{21100.8 \: cm^2}}$

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• Given:-

• Cylindrical tank has radius of 80cm

• Depth of the cylindrical tank is 2cn

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• To Find:-

• Area of Wetted surface

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• Solution:-

We need to find the Total Surface Area of the cylindrical tank.

Hence,

T.S.A of cylindrical tank = Area of base of tank + C.S.A of tank

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Therefore,

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Area of wetted surface will be

$\purple{\bigstar}$ $\large\underline{\boxed{\bf\blue{T.S.A = \pi r^2 + 2 \pi rh}}}$

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➪ $\sf T.S.A = 3.14 \times (80)^2 + 2 \times 3.14 \times 80 \times 2$

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➪ $\sf T.S.A = 3.14 \times 6400 + 6.28 \times 160$

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➪ $\sf T.S.A = 20096 + 1004.8$

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★ $\large{\bf\pink{T.S.A = 21100.8 \: cm^2}}$

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Therefore, the area of wetted surface is 21100.8 cm²

Answer 2

Answer:

\red{\bigstar}★ Area of wetted surface \large\leadsto\boxed{\tt\green{21100.8 \: cm^2}}⇝

21100.8cm

2

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• Given:-

Cylindrical tank has radius of 80cm

Depth of the cylindrical tank is 2cn

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• To Find:-

Area of Wetted surface

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• Solution:-

We need to find the Total Surface Area of the cylindrical tank.

Hence,

T.S.A of cylindrical tank = Area of base of tank + C.S.A of tank

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Therefore,

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Area of wetted surface will be

\purple{\bigstar}★ \large\underline{\boxed{\bf\blue{T.S.A = \pi r^2 + 2 \pi rh}}}

T.S.A=πr

2

+2πrh

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➪ \sf T.S.A = 3.14 \times (80)^2 + 2 \times 3.14 \times 80 \times 2T.S.A=3.14×(80)

2

+2×3.14×80×2

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➪ \sf T.S.A = 3.14 \times 6400 + 6.28 \times 160T.S.A=3.14×6400+6.28×160

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➪ \sf T.S.A = 20096 + 1004.8T.S.A=20096+1004.8

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★ \large{\bf\pink{T.S.A = 21100.8 \: cm^2}}T.S.A=21100.8cm

2

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Therefore, the area of wetted surface is 21100.8 cm²