Question

9. 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

Answer:

Given, 2r=7

⇒r=7/2;h=4

Area of wetted surface = CSA of cylinder ABCD + Area of base (Circle)

=2πrh=πr

2

=πr(2h+r)

=

7

22

×

2

7

(8+3.5)

=11×11.5=126.5 sq. m

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

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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