A drop of mercury of radius 3mm splits into 27 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 N/m.)

Let r be the radius of small droplet. As the volume of mercury will remain constant on splitting the drop into droplets, therefore


Let r be the radius of small droplet. As the volume of mercury will r
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Calculate the force required to take away a flat circular plate of radius 5cm and mass 2gm from the surface of water. The surface tension of water is 72 dyne/cm.

Given, 
Radius of the circular plate, r = 5 cm 
Mass, m = 2 gm
Surface tension of water, T = 72 dyne/cm

 Weight of plate is, W = mg=2 x 980 = 1960 dyne

The circumference of circular plate is given by, =2πr=2π×5=31.4cm

The down pull on plate due to surface tension is given by, 
 
f=T=72×31.4=2260.8 dyne 

Therefore, total force required to take plate away from water surface is,

F=W+f 

 =1960+2260.8 

 =4220.8 dyne

 

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Water rises to a height 4.8cm in a certain capillary tube. When the same tube is dipped in some another liquid it falls by 12.5cm. Compare the surface tension of water and liquid. Density of liquid is 9.6 gm/cc and angle of contact of liquid is 120°.

Given, 

Rise in water = 4.8 cm 
Fall in water when tube is dipped in some other liquid = 12.5 cm
Density of liquid = 9.6 gm/cc
Angle of contact of liquid = 120o

The height to which the liquid rises or falls in the capillary tube is given by the relation,

                        h=2T cosθrpg 

The above relation can be simplified to 

                        T=rgph2 cos θ                    ....(1)

Let T1 and T2 be the surface tension of water and liquid respectively.

By using the equation (1), we have


T1T2=ρ1h1ρ2h2cosθ2cosθ1

Here, ρ1=1 gm/cc,        h1=4.8cm,        θ1=0°         ρ2=9.6gm/cc,      h2=-12.5cm,   θ2=120°    T1T2=1×4.89.6×(-1.25)cos 120°cos 0°=15        

 
 
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A big drop of water is sprayed into 1000 small droplets. Find the ratio of surface potential energy of drop to single droplet?


Let r and R be the radius of small droplet and big drop. As the volume of water will remain constant on splitting the drop into droplets, therefore


Let r and R be the radius of small droplet and big drop. As the volum
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Why mercury does not cling to glass?

The force of cohesion between mercury molecules is much greater than force of adhesion between mercury and glass. Therefore, mercury does not cling to glass. 


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