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A cylindrical tank is open at top of diameter 0.18 m and height 0.9 m. Water is filled in it up to a height of 0.18 m . How much time will be taken to

empty the tank through the small hole of radius 0.0003 m? g=10 m/s2.



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I suppose there is a small hole at the bottom planar surface of the cylindrical tank.
      radius of the tank = R = 0.09 m
       height of water inthe tank = H = 0.18 m
     Volume of water in the tank = V = π R² H =  π * 0.09² * 0.18  m³
                               = 0.001458 π  m³

     radius of small hole = r = 0.0003 m
    area of cross section of the hole = A = π r² = * 0.0003² = 9 π * 10⁻⁸  m²

Let speed of water coming out of small hole = v
   As the mechanical gravitational energy is conserved, the PE of water at the top of the water surface is converted in to the kinetic energy of the water coming out of the hole.
                      d ρ g h = 1/2 d ρ v²
                       v = √ (2 g h)
  As the water level decreases, the speed of water coming out of the hole reduces.  So the the time rate of water outflow is variable.  we need to do integration for solving this exercise.

    Volume of water in the tank = V = π R² h
    Volume of water coming out of the small hole in Δt time duration
            ΔV =  A * v * Δt
       =>  - π R² Δh  = π r² * v * Δt
             we put a minus sign because,  h decreases as t increases.

     =>  dh/dt = - v  r²/R²   = [ - √(2g) r²/R² ] √h
     =>  dh / √h  = - √(2g) r²/R²  dt
  Integrating with limits as  h = H at t = 0,  and  h = 0, at  t = T, we get

     =>  2 * √H  -  0  =  √(2g) r²/R² * T

   =>  T =  R²/r²  * √(2H/g)
Time\ duration\ to\ empty\ tank=\frac{R^2}{r^2}*\sqrt{\frac{2H}{g}}\\\\So\ T = \frac{0.09^2}{0.0003^2}\sqrt{\frac{2*0.18}{10}}=17076 sec

so Time duration =  4 hours 44 minutes  36.3 seconds

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