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1)a water tank which is on ground has an arrangement to maintain a constant water levvel of depth 60cm.thru a hole on its vertical wall at a depth of 20cm

from the free surface water comes out and reaches the ground at a certain distance.to have the same horizontal range anothe r hole can be made at a depth of? 2)a tank full of water has a small at its bottom.if one fourth of the tank is emptied in t1 sec and the remaining three-fourths of the tank is emptied in t2 sec.the ratio t1/t2= 3)a large tank filled with water to a height h is to be emptied thru a small hole at the bottom.the ratio of the time taken for the level to fall from h to h/2 and that taken for the level to fall from h/2 to 0 is

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Bernoulli's energy conservation principle for the fluids tells the relation between pressures and kinetic energy at two locations in a fluid.

Total Energy at the surface of water = Total energy at the hole at 20 cm depth P₁ + 1/2 ρ v₁² + ρ g h₁ = P₂ + 1/2 ρ v₂² + ρ g h₂

P₁ = P₂ = atmospheric pressure, g = 10 m/sec² v₁ = 0 as the water at the surface is maintained stationary. h₁ - h₂ = h = 20 cm = 0.20 m

So v₂ = √ (2gh) --- equation 1

Hence, v₂² = 2 * 0.20 g = 4 => v₂ = 2 m/sec

We find the range for the water particles with this speed horizontally directed and at a height of 0.40 meters above ground.

t = time to reach ground = √ (2 * 0.40 / g ) = 0.2 *√2 sec Range = v₂ t = 0.4 √2 meters

If h is the depth of hole from water surface, then v = √(2gh) time to reach ground = t = √ [2*(0.60-h)/g ] Then range R = v₂ t = 2 √[ h (60-h) ] = 0.4√2 --- equation 2

Range will be equal for h = 20cm, and h = 40 cm, as the product of h and (60-h) will be same. If you solve quadratic equation 2, we get the same. ============================================== 2 ) velocity v₂ of water coming out of hole at the bottom of tank = √(2 g h) from equation 1 above. Here we neglect v₁ at the water surface as it is very small as compared to v₂. Let A be the area of cross section of the tank.

Volume of water in tank = V = A h water flow rate out of tank = decrease rate in volume inside tank A v₂ = d V/dt = - A dh/dt => v₂ = √(2 g h ) = - dh/dt

=> - dh/√h = √(2g) dt ------- equation 3 Integrating the expressions on both sides ,

Time for water level to go from h₀ to 3 h₀/4, is

Time for water level to go from 3 h₀ /4 to 0 is

Ratio t1/t2 =

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3) Using equation 4, above we get the values for t1 and t2.