A huge tank is filled with some fluid and is covered (not open to the atmosphere

Question

A huge tank is filled with some fluid and is covered (not open to the atmosphere). A small hole is punched 32.1 meters below the top of the fluid line on the top of the tank. The pressure on the fluid at the top is a vacuum because there isn't much air between the fluid line and the top of the tank. At the small hole, the pressure is 84.6 percent of one atmosphere because this tank is not at sea level. At the hole, the water leaves at a velocity of 12 m/s. This fluid is then poured into a pool. A log of Balsa wood with a density of 160 kg/m3 is placed in this fluid. By volume, what percentage of the log above the fluid line?

Explanation / Answer

Let rho is the density of the fluid.

Apply Bernoulli's equation

P1 + rho*g*h1 + 0.5*rho*v1^2= P2 + rho*g*h2 + 0.5*rho*v2^2

here, P1 = 0, V1 = 0

0 + rho*g*h1 + 0 = P2 + rho*g*h2 + 0.5*rho*v2^2

rho*g*(h1-h2) = P2 + 0.5*rho*v2^2

rho*9.8*32.1 = 0.846*1.013*10^5 + 0.5*rho*12^2

314.58*rho = 85700 + 72*rho

rho*(314.58 - 72) = 85700

rho = 85700/(314.58 - 72)

= 353 kg/m^3

now Let V is the volume of the log.

let m is the of the log

when the log is in equilibrium, net force acting on it = 0

so, Fnet = 0

B - m*g = 0 (here B is buoynat force)

weight of dispalced fliud - m*g = 0

rho_fluid*V_inside*g = m*g

rho_fliud*V_inside*g = rho_log*V*g

V_inside/V = rho_log/rho_fluid

= 160/353

V_inside = 0.453*V

so, V_outside = V - V_insde

= V - 0.453*V

= 0.547*V

= 54.7 % of V <<<<<<-----------Answer

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