# (i)A mettalic sphere floats in an immiscible mixture of water (pw=10^3)and a liquid (pl=13.5*10^3 such that 4/5 portion in water and remaining in liquid.The density of metal is (ii)A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8The relative density of siver is 10.5.The tension in string is (iii)the ratio of diameters of certain air bubble at bottom and at6 surface is 1 is to 2.What is depth of lake(1 atm = 10 m depth of water) please answer all with detailed explanations

1
by rajusetu

2014-12-21T03:19:04+05:30

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1)
4/5 * V * 10^3 * g + 1/5 * V * 13.5 * 10^3 * g = density * V * g

density = 3.5  * 10^3  kg/cubic meter

2)
Tension = weight - buoyancy force
= 2.1 * g -  (2.1/10.5) * 0.8 * g
= 1.94 * g  newtons

3 )    If you do not know surface tension, please look at the simplified solution.

Pressure of air in side bubble * volume = constant at constant temperature
[ P_atm + density of water * g * h ] * 4/3 π R³ = p_atm * 4/3 π (2R)³

Density * g * h = 7 * P_atm
= 7 * density * g * 10 meters  (given)
h = 70 meters

=============================================
Let S be surface tension., energy per unit surface area.

Bubble at surface of water :
Pressure of air inside P1 = 2 S / 2 R + P_atm
volume of bubble V1 = 4/3*π (2R)³

bubble at a depth  h
Pressure P2 = 2 S / R + P_atm + 10^3 * g * h
volume of bubble V2 =  4/3 π R³

P1 V1 = P2 V2 at constant temperature
P1 * 8 *V2 = P2 * V2

P2 = 8 P1  =
2 S / R + P_atm + 1000 * g h = 8 (S / R + P_atm)

1000 * g * h = 6 S / R + 7 P_atm

Let us ignore the surface tension part.  It may not be significant.

1000 * g h = 7 * 1000 * g * 10 meters

h = 70 meters