Find the excess of pressure inside the soap bubble of diameter 1.2mm. The surface tension of soap solution is 18 dyne/cm.

Given, 
Diameter of the soap bubble, d = 1.2 mm 

Surface tension of soap, T = 18 dyne/cm

So, excess of pressure inside the soap bubble is given by, 


                                            Δp=4TrHere, T=18 dyne/cm           D=1.2mm=0.12cm Excess pressure, p =4x180.06                                             =1.2×103 dyne/cm2                                              =1.2×102 N/m2 
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A 10x1.5x0.2 cm3 glass plate weighs 8.6 gm in air. Now it is immersed half in water with longest side vertical. What will be its apparent weight? (surface tension of water is 70 dyne /cm.)

The different forces acting on the plate are:

(i) Weight W vertically downward,
(ii) Upward thrust U vertically upward,
(iii) Surface tension force T in downward direction .

These forces are given by, 

Weight, W=8.6 x 980 = 8428 dyne

Upward Thrust, U=weight of water displaced 

                          =Volume of water displaced x density of water x g
                           =12(10x1.5x0.2)x1 x 980=1470 dyne

Let,

T = Total length of the water in touch with plate x surface tension

   =2(1.5 + 0.2) x 70 = 238  dyne

Now the apparent weight, 

W+ W + T - U = 8428 + 238 - 1470

                         = 7196 dyne

                         = 7.434 gf

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What will be the pressure inside a small air bubble of 0.1 mm radius situated just below the free surface of water?

(S.T. of water = 72dyne/cm; Atmospheric pressure 1.013 x 106 dyne/cm2 )


We know that excess of pressure inside the air bubble is,


We know that excess of pressure inside the air bubble is,
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A vessel, whose bottom has a hole of diameter 0.2mm, is filled with water. Find the maximum height to which the water can be filled without leaking it from the hole. (Surface tension of water is 70 dyne/cm)

Let the vessel be filled to a height of h.

The force exerted by height column of water will be balanced by surface tension force on the free surface of water at the hole, at equilibrium.

That is,

        ρgh×πr2=2πr×T                  h=2TrpgHere,   T= 70 dyne/cm              r=d/2=0.1mm=0.01 cm,             ρ=1 gm/cc,              g= 1000 cm/s2   h=2×700.01×1×1000=14cm 

So, 'h' is the maximum height to which the water can be filled without leaking it from the hole. 
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The excess of pressure inside the bubble of radius 2-4 cm is balanced by 0-12 cm height column of oil of density 0.8 gm/cc. Find the surface tension of liquid.


Given, the excess of pressure inside the bubble is, height column of oil of density 0.8 gm/cc. 

Find: surface tension of liquid?

The formula for excess pressure is given by, 

                   Δp=4Tr

The pressure at height column of oil is given by, 

                   Pm = pgh

Now, excess of pressure inside the bubble is balanced by height column of oil.

Therefore, 

                      4Tr=ρgh                     T = ρrgh4Here       r=2.4 cm, ρ=0.8 gm/cc               h=0.12 cnm, g=980 cm/s2  Surface tension of the liquid, T = 0.8×2.4×980×0.124                                                             = 56.45 dyne/cm
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