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 , 

2\ [-\sqrt{h}]_{h_0}^{h_t} = \sqrt2\ g\ t\\\\ t=\frac{\sqrt2}{g}[ \sqrt{h_0}-\sqrt{h_t} ],\ \ \ -- equation\ 4\\

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

t_1=\frac{\sqrt2}{g}\ [\sqrt{h_0}-\frac{\sqrt{3}}{2}\sqrt{h_0}\ ]=\frac{\sqrt{2h_0}}{2g}[ 2-\sqrt3 ]\\\\

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

t_2=\frac{\sqrt2}{g}\ [\frac{\sqrt{3}}{2}\sqrt{h_0}-0\ ]=\frac{\sqrt{6h_0}}{2g}\\

Ratio t1/t2 = 

\frac{t_1}{t_2}= \frac{2-\sqrt3}{\sqrt3}\\


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

t_1=\frac{\sqrt2}{g}[ \sqrt{h_0} - \sqrt{\frac{h_0}{2}} ]=\frac{(\sqrt{2}-1)\sqrt{h_0}}{g}\\\\t_2=\frac{\sqrt{h_0}}{g}\\\\Ratio\ t_1/t_2 = \sqrt{2}-1\\

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