Answers

2015-05-18T14:46:12+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

1 5 1
please click on thanks blue button above