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Asked by sujalsinhaofficial07 | 15 Jan, 2023, 19:50: PM
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)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/6269d0bb37d4fecc110442b18d8f84cb63c4345a8dad64.71050011f3.png)
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](https://images.topperlearning.com/topper/tinymce/cache/b58a671b63b78e27e53a1a30be70d9df.png)
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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](https://images.topperlearning.com/topper/tinymce/cache/e32b0b2b7a3254fb52e0ec753f25c59f.png)
τ = 1.333 × 104 N m
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