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A vertical rectangular plane surface is submerged in water such that its top and bottom surfaces are 1.5 m and 6.0 m respectively below the free surface. The position of centre of pressure below the free surface will be at a distance of
  • a)
    3.75 m
  • b)
    4,0 m
  • c)
    4.2 m
  • d)
    4.5 m
Correct answer is option 'C'. Can you explain this answer?
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Given data:
Top surface of the rectangular plane = 1.5 m below the free surface
Bottom surface of the rectangular plane = 6.0 m below the free surface

To find: distance of the centre of pressure below the free surface

Formula:
The distance of the centre of pressure below the free surface is given by the formula:

zp = (Ic/As) + zcg

where,
zp = distance of the centre of pressure below the free surface
Ic = moment of inertia of the submerged area
As = area of the submerged portion of the rectangular plane
zcg = distance of the centroid of the submerged portion of the rectangular plane from the free surface

Calculation:
The area of the submerged portion of the rectangular plane can be calculated as:

As = length × breadth × depth
As = 1 × 1.5 × (6 - 1.5)
As = 7.5 m²

The centroid of the submerged portion of the rectangular plane will be at a distance of:

zcg = (1.5 + 6)/2
zcg = 3.75 m

The moment of inertia of the submerged area can be calculated as:

Ic = (1/12) × breadth × depth³
Ic = (1/12) × 1.5 × (6 - 1.5)³
Ic = 1.71875 m⁴

Substituting the values in the formula, we get:

zp = (Ic/As) + zcg
zp = (1.71875/7.5) + 3.75
zp = 4.2 m

Therefore, the distance of the centre of pressure below the free surface is 4.2 m.

Hence, option (c) is the correct answer.
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A vertical rectangular plane surface is submerged in water such that its top and bottom surfaces are 1.5 m and 6.0 m respectively below the free surface. The position of centre of pressure below the free surface will be at a distance ofa)3.75 mb)4,0 mc)4.2 md)4.5 mCorrect answer is option 'C'. Can you explain this answer?
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