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 A circular opening, 6m diameter, in a vertical side of a tank is closed by a disc of 6m diameter which can rotate about a horizontal diameter. Calculate the force on the disc. The centre of circular opening is at the depth of 5 m.
  • a)
    1.38 MN
  • b)
    2.76 MN
  • c)
    5.54 MN
  • d)
    7.85 MN
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A circular opening, 6m diameter, in a vertical side of a tank is close...
Explanation: F=w*A*ŷ
=9.81*1000*3.142*32*5
=1.38 MN.
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Most Upvoted Answer
A circular opening, 6m diameter, in a vertical side of a tank is close...
Given:
Diameter of circular opening (d) = 6m
Depth of center of the opening from the surface (h) = 5m

To find:
Force on the disc

Solution:
The force on the disc can be calculated using the following formula:

F = γhA

where,
γ = unit weight of the liquid (in N/m³)
h = depth of the center of the opening from the surface (in m)
A = area of the opening (in m²)

Area of the opening:
A = πd²/4
A = π(6)²/4
A = 28.27 m²

Unit weight of water:
γ = 9.81 kN/m³

Converting kN to N:
γ = 9810 N/m³

Substituting the values in the formula:

F = 9810 x 5 x 28.27
F = 1,386,667 N

Converting N to MN:
F = 1.38 MN

Therefore, the force on the disc is 1.38 MN (Option A).
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A circular opening, 6m diameter, in a vertical side of a tank is closed by a disc of 6m diameter which can rotate about a horizontal diameter. Calculate the force on the disc. The centre of circular opening is at the depth of 5 m.a)1.38 MNb)2.76 MNc)5.54 MNd)7.85 MNCorrect answer is option 'A'. Can you explain this answer?
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