Question

In a container with a circular base (R = 0.2 m) water (density ρ = 997 filled up to a height h 0.5m Hint:Assume a frictionless fluid. Note that R. ) is Im Figure 2: water container a) At which height from the bottom of the container an opening should be put so that the water hits the ground at the furthest distance from the jar? (2P) b) Now the opening (circular, r0.5 ) is put at y 0.3m. Calculate how the water level h(t) changes over time. (3P)

Answer 1

h = 05 m R = 0.2 m rho = 997 kg/m^3 (a) Bernoulli's Principle at 1 & 2. P_1 + 1/2 rho V_1^2 + rho gh_1 = P_2 + 1/2 rho V_2^2 + rho gh_2 P_1 = P_2 = P_0 V_1 = V_R V_2 = V_r h_1 = h h_2 = y P_0 + 1/2 rho V_R^2 + rho g h = P_0 + 1/2 rho V_r^2 + s g y A_R V_R = A_r B_r A_R = pi R^2/4 pi R^2/4 V_r = pi r^2 V_r A_r = pi r^2 R > > r V_R lessthanorequalto V_r V_R^2 is neglected V_r = squareroot 2g (h - y) \r\n V_r = squareroot 2g (h - y) V_r = squareroot 2g (h - y) Along y-direction time taken -y = -1/2 gT^2 T = squareroot 2y/y X = V_r T X = V_r squareroot 2y/g X = squareroot 2g (h - y) squareroot 2y/g X = squareroot 2 (h - y) y X is maximum X^2 = Z = 2y (h - y)