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Asked by sujalsinhaofficial07 | 15 Jan, 2023, 19:50: PM

Expert Answer

Part (a)

Force on bottom surface = Pressure × area

Pressure = ( ρ × g × d ) = (10³ × 10 × 1) Pa = 10⁴ Pa

where ρ is density of water , g is acceleration due to gravity and d is depth of water

Area = ( 3 × 2 ) m² = 6 m²

Force on bottom surface = 10⁴ Pa × 6 m² = 6 × 10⁴ N

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Part (b)

Pressure at a depth x = ρ × g × x

Force dF on strip of width dx and length 2 m = pressure × area

dF = ( ρ × g × x ) × ( 2 × dx ) = ( 2 ρ g ) ( x dx )

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Part (c)

Torque dτ = force × distance

Torque dτ = ( 2 ρ g ) ( x dx ) × (1-x)

Torque dτ = ( 2 ρ g ) [ ( x - x² ) dx ]

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Part (d)

Total force F on side of dimension ( 2 m × 1 m ) is

begin mathsize 14px style F space equals space integral d F space equals space 2 space rho space g integral subscript 0 superscript 1 x space d x space equals space rho g space equals space 10 cubed cross times 10 space N space equals space 10 to the power of 4 to the power of space end exponent N end style

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Part (e)

Total torque τ is

begin mathsize 14px style tau space equals space integral d tau space equals space 2 space rho space g integral subscript 0 superscript 1 left parenthesis space x space minus space x squared right parenthesis d x space equals space 4 over 3 space rho space g end style

τ = 1.333 × 10⁴ N m

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Answered by Thiyagarajan K | 15 Jan, 2023, 22:59: PM

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