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If two liquids of same volume but different densities ρ1 and ρ2 are mixed, then density of mixture is given by





(a) ρ=Total massTotal volume=m1+m22V=V(ρ1+ρ2)2V=ρ1+ρ22

With rise in temperature, density of a given body changes according to one of the following relations





(b) Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease. 

i.e. ρρ0=m/Vm/V0=V0V=VV0(1+γθ)=(1-γθ) ρ=ρ0(1-γ θ)

An ice berg of density 900 Kg/m3 is floating in water of density 1000 Kg/m3. The percentage of volume of ice-cube outside the water is 





(c) Let the total volume of ice-berg is V and its density is ρ. If this ice-berg floats in water with volume Vin 

inside it then Vinσg=VρgVin=ρσV

or Vout=V-Vin=σ-ρσVVoutV=σ-ρσ=1000-9001000=110 Vout= 10% of V

A hemispherical bowl just floats without sinking in a liquid of density 1.2×103 kg/m3. If outer diameter and the density of the bowl are 1 m and 2×104 kg/m3 respectively, then the inner diameter of the bowl will be 





(c) Weight of the bowl = mg

=Vρg=23πD23-d23ρg

where D = Outer diameter ,

d = Inner diameter

ρ = Density of bowl                               

Weight of the liquid displaced by the bowl

=Vσg=23πD23σg

where σ is the density of the liquid

For the floatation    23πD22σg=23πD23-d23ρg123×1.2×103=123-d232×104

By solving we get d = 0.98 m.

A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be 





(b) Let specific gravities of concrete and saw dust  are ρ1 and ρ2 respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere

43π(R3-r3)ρ1g+43πr3ρ2g=43πR3×1×gR3ρ1-r3ρ1 +r3ρ2=R3R3(ρ1-1)=r3(ρ1-ρ2)R3r3=ρ1-ρ2ρ1-1R3-r3r3=ρ1-ρ2-ρ1+1ρ1-1(R3-r3)ρ1r3ρ2=1-ρ2ρ1-1ρ1ρ2Mass of concreteMass of saw dust=1-0.32.4-1×2.40.3=4

A metallic block of density 5 gm cm-3 and having dimensions 5 cm × 5 cm × 5 cm is weighed in water. Its apparent weight will be





(d) Apparent weight 

=V(ρ-σ)g =l×b×h×(5-1)×g=5×5×5×4×g Dyne = 4×5×5×5 gf

A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kg-wt is





(b) Apparent weight 

=V(ρ-σ)g=Mρ(ρ-σ)g

=M1-σρg=2.11-0.810.5g=1.94g N=1.94 kg-wt

Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with 2/3 of its volume immersed. Compare the densities of A and B





(c) If two different bodies A and B are floating in the same liquid then

ρAρB=(fin)A(fin)B=1/22/3=34

The fraction of a floating object of volume V0 and density d0 above the surface of a liquid of density d will be





(c) For the floatation  V0d0g=Vindg

Vin=V0d0d

 Vout=V0-Vin=V0-V0d0d=V0d-d0dVoutV0=d-d0d

Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. This law was first formulated by:






(d)

According to Pascal's principle, the pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid with the walls of the containing vessel too. A common application of this principle is in hydraulic lift, to lift the car. Here, a small amount of force applied to the small-area piston, which will be transformed into a large force at a large-area piston.

A cork is submerged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevator moving with acceleration downwards, the length of spring





(b)

Upthrust reduces from Vdg to Vdg-a which is responsible for the extension in the spring.

A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water ?





(a) When body (sphere) is half immersed, then
upthrust = weight of sphere 

V2×ρliq×g=V×ρ×g     ρ=ρliq2

When body (sphere) is fully immersed then,

Upthrust = wt. of sphere + wt. of water poured in sphere

V×ρliq×g=V×ρ×g + V'×ρliq×gV×ρliq=V×ρliq2+V'×ρliqV'=V2

A ball whose density is 0.4×103 kg/m3 falls into water from a height of 9 cm . To what depth does the ball sink





(b) The velocity of ball before entering the water surface

v=2gh=2g×9

When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)

The retardation, a=apparent weightmass of ball

=V(ρ-σ)g=ρ-σρg=0.4-10.4×g=-32g

If h be the depth upto which ball sink, then,

0-v2=2×-32g×h2g×9=3gh h=6 cm

 

A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water ?





(c)

ρball × Vball × g = ρwater × Vdisplaced × gAs ρball  >  ρwaterVdisplaced > VballSo water level falls

Two pieces of metal when immersed in a liquid have equal upthrust on them; then





(c) Since, upthrust (F) =Vσg   i.e. F∝V