(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
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kvnmurty:
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
Answered by
51
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
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
ok
yes sir
Answered by
10
Answer:
7H
Explanation:
P1V1=P2V2 (1 −> bottom of lake 2−> surface of lake)
(h+H)(43πr13)=H(43πr23)(h+H)(r13)
=H(r23)h+HH=(r2r1)3hH+1=(2r1r1)3hH=8−1h=7H
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