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A water tank which is on ground has an arrangement to maintain a constant water level of depth 60cm. Through 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 another hole can be made at a depth of?

A water tank which is on ground has an arrangement to maintain a constant water level of depth 60cm. Through 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 another hole can be made at a depth of?

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
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.

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