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Asked by sujalsinhaofficial07 | 15 Jan, 2023, 07:50: 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)

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

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Part (e)

Total torque τ is

τ = 1.333 × 104 N m

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Answered by Thiyagarajan K | 15 Jan, 2023, 10:59: PM
JEE main - Physics