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Bernoulli's theorem is based on :


Two spherical soap bubbles of radii r1 and r2 in vaccum collapse under isothermal condition. The resulting bubble has radius equal to :

4for isothermal conditionPV = P1V1 + P2V24Tr43πr3 = 4Tr143πr13 + 4Tr2.43πr23r2 = r12 + r22

The energy needed in breaking of a drop of liquid of radius R into n drops of radius r is given by (T is surface tension and P is atmospheric pressure) :

Surface area of large sphere=4πR2Surface area of n small spheres=n4πr2Work done=T×A=T×n4πr2-4πR2


A rectangular film of liquid is extended from (4 cm× 2 cm) to 5 cm×4×cm. If the work done is 3×10-4 J,the value of the surface tension of the liquid is 


(b) Increases in surface energy= work done in area×Surface tension 

Increase in surface area,

 A=5×4-4×2×2                       flim has two surfaces

=20-8×2 cm2=24 cm2=24×10-4 m2 

So, work done, W=T.A

  3×10-4=T×24×10-4 T=18=0.125 N/m


Three liquids of densities ρ1, ρ2 and  ρ3 (with p1>p2>p3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ1,θ2 and θ3 obey


(b) According to ascent formula for capillary tube,

              h=2Tcos θpgr cosθ1p1=cosθ2p2=cosθ3p3Thus,cosθρ p1>p2>p3 cos θ1>cosθ2>cos θ3   0θ1<θ2<θ3<π2


Two non-mixing liquids of densities ρ and nρ(n>1) are put in a container. The height of each liquid is h. A solid cylinder  floats with its axis vertical and length pL p<1 in the denser liquid. The density of the cylinder is d. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to 

According to Archimedes' principle, the buoyant force on the cylinder is equal to the weight of the liquid displaced by the cylinder. The volume of the liquid displaced is pLπr^2, where r is the radius of the cylinder. The weight of this liquid is pLπr^2ρn. This weight must equal the weight of the cylinder, which is pLπr^2d. Therefore, d = ρn.

The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10-11 Pa-1 and density of water is 103kg/m3. What fractional compression of water will be obtained at the bottom of the ocean?

Given d=2700 m


Compressibility=45.4 x 10-11 Pa-1  per pascal

The pressure at the bottom of ocean is given by


So, fractional compression=compressibility x pressure

=45.4 x 10-11x 27 x106


A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (Pair=1.2kg/m3)

From Bernoulli's theorem


where, p1 and p2 are pressure inside and outside the roof and v1 v2 are velocities of wind inside and outside the roof, Neglect the width of the roof.

Pressure difference is



Force acting on the roof is given by

F=(p1-p2)A=960 x 250=24x104N=24X105N

As the pressure inside the roof is more than outside to it. So the force will act m the upward direction, i.e. F=24x105N,upward

The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is

The liquid flow is incompressible and steady. By the principle of conservation of mass, the volume flow rate must be constant throughout the tube. Therefore, the speed v in the tube is related to the speed v' through the holes by the equation πR^2v = nπr^2v', which gives v' = vR^2/nr^2.

The heart of a man pumps 5L of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6 x 103 kg/m3 and g =10 m/s2, then the power of heart in watt is

Given pressure=150mm of Hg

Pumping rate of heart of a man=dVdt=5x10360 m3/s

Power of heart=P.dVdt=ρgh.dVdt [P=ρgh]

=>(13.6x103kg/m3) (10x0.15x5x10-3)/60


Water rises to a height h in capillary tube . If the length of capillary tube is above the surface of water is made less than h, then

In case of insufficient length, angle of contact adjusts itself so that water does not spill

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

If the surface area changes, it will change the surface energy as well. As the surface area is decreasing, energy will be released.Change in surface energy=T×AIf there is n no. if small drops;Volume of a large drop=volume of n small drops43πR3=n×43πr3Change in area;A=4πR2-n×4πr2=34πR33R-n×4πr33r=34πR33R-4πR33r=3VR-Vr=3V1R-1rE=T×A=3VT1R-1r, released.

The wettability of a surface by a liquid depends primarily on

(d) The value of angle of contact determines whether a liquid will spread on the surface

An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

Let m is the mass per unit length then rate of mass per sec=mxt=mv

Rate of KE=12mvv2=12mv3

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm3. If the mass of the other is 48 g, its density in g / cm3 is 

(c) Apparent weight = V(ρ-σ)g=mρ(ρ-σ)g
where m = mass of the body,
ρ = density of the body
σ = density of water
If two bodies are in equilibrium then their apparent weight must be equal.


By solving we get ρ2=3.