A cylindrical tank 7cm in diameter contains water to a depth of 4m. What is the total area of wetted surface
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Answered by
12
hello....
▶▶surface [wet] = curved surface + bottom circle
▶▶radius of the cylinder r = 7/2
▶▶height of the cylinder h = 4
▶▶Area of wet surface = CSA of cylinder + Ar base circle = 2πrh + πr^2
⇒ πr(2h + r)
⇒22 x 7(8 + 3.5)7 x 2
⇒ 11 x 11.5
= 126.5 m^2
thank you...
▶▶surface [wet] = curved surface + bottom circle
▶▶radius of the cylinder r = 7/2
▶▶height of the cylinder h = 4
▶▶Area of wet surface = CSA of cylinder + Ar base circle = 2πrh + πr^2
⇒ πr(2h + r)
⇒22 x 7(8 + 3.5)7 x 2
⇒ 11 x 11.5
= 126.5 m^2
thank you...
Answered by
7
Given : d = 7m , depth = 4 m
Wet area = CSA under water + area of base
Wet area = πdh + π(d/2)^2
= 3.14×7×4 + 3.14×(7/2)^2 cm^2
= 87.92 + 38.465 cm^2
= 126.385 cm^2
So, the total area of the wet surface
= 126.385 cm^2.
please mark as BRAINliest answer
Wet area = CSA under water + area of base
Wet area = πdh + π(d/2)^2
= 3.14×7×4 + 3.14×(7/2)^2 cm^2
= 87.92 + 38.465 cm^2
= 126.385 cm^2
So, the total area of the wet surface
= 126.385 cm^2.
please mark as BRAINliest answer
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