(i)when a large bubble rises from the bottom of a lake to the surface its radius doubles.The atm =that of a column of water of heigth H.the depth of the lake is
(ii)A motor ship sails from sea water to a river.To keep the same draugth a 90 tonne load is removed from the ship>find the mass of the loaded ship before it has been unloaded
(iii)A cube with an edge of 10 cm is immersed in a vessel containing water.A layer of liquid immiscible with water and having a density of 0.8*10^3 kg/m^3.is poured above water.The interface betwwen the liquid is at the middle of the cube heigth.The mass of the cube is
PLease answer all questions with detailed explanations

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why are u creating multi part questions? please create only one part - one question -- it is difficult to answer and it is difficult to moderate.
why are u creating multi part questions? please create only one part - one question -- it is difficult to answer and it is difficult to moderate
sorry

Answers

2014-12-22T07:36:37+05:30

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Volume of the bubble is inversely proportional to the pressure of air inside bubble.
               Pressure * volume = constant at same temperature.

   As the bubble raises to the top, the pressure on the bubble by water at that level decreases and hence bubble expands.

  Bubble's volume is 8 times at surface compared to volume at the bottom of lake.  So pressure at the bottom will be 8 times that at the surface.  Reason is the radius is twice and hence volume is 2^3 times.
 
  Pressure at surface of water lake =  1 atm = water density * H * g  (given)
  Pressure at the bottom of lake = 8 * 1 atm = 8 atm  = density * 8 * H * g
                           = density * depth * g + 1 atm at the surface 
         Hence the depth of lake = 8 * H - 1 * H = 7 * H

2)
        density of sea water = d_sea  = 1030 kg/m³
          density of river water = d_river  = 1000 kg/m³
        total mass of loaded ship = M tonnes
        let  immersion depth = H

          d_sea * H * Area * g = M * g
          d_river * H * Area * g =  (M - 90) * g
             =>      90 tonnes = (d_Sea - d_river) H * Area
             M = 90 * d_sea / (d_Sea  - d_river)    tonnes
       calculating, M = 90 * 1030 / (1030 - 1000) = 3090 tonnes

3)
           mass of cube = M
        volume of cube immersed in liquid = 1/2 * 0.10³ = 0.0005 m³
        volume of cube immersed in water = 0.0005 m³

       0.0005 * 1000 * g + 0.0005 * 800 * g = M g

             M = 0.9 kg

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