Consider the wall of a dam to be straight with height H and length L . It holds a lake of water of height h h < H on one side. Let the density of water be ρ w Denote the torque about the axis along the bottom length of the wall by τ 1 . Denote also a similar torque due to the water up to height h / 2 and wall length L / 2 by τ 2 . Then, τ 1 / τ 2 (ignore atmospheric pressure) is

Step-by-step solution

Consider a strip of width d y at a depth h - y below the surface. Torque on bottom length ( O Z in diagram) due to force of water on strip of width d y is d τ = Force × Perpendicular distance = Pressure × Area × Distance d τ = ρ g h - y · L · d y · y So, torque on bottom length due to complete volume of water, τ 1 = ∫ 0 h L ρ g h - y y · d y = L ρ g ∫ 0 h h y - y 2 d y = L ρ g h y 2 2 - y 3 3 0 h = L ρ g h 3 2 - h 3 3 or τ 1 = L ρ g h 3 6 . . . ( i ) and torque due to water upto height h 2 and wall length 1 2 , τ 2