Question

Calculate the amount of energy evolved when 8 drops of mercury of radius 1mm each ,combine to form 1drop (surface tension of Hg = .55/m)

Answer 1

Answer:

Here, 

T=525×10−3Nm−1

R=1×10−3

Now, the surface area of the bigger drop.

=4πR2

=4×3.14×(1×10−3)2

then, 

S1=0.1256×10−4m2

Now, the total initial volume =34πR3

and the volume smaller drops =100034πR3

consider r be the radius of smaller drops,

Hence, volume of droplets =34πr3 

and 34πr3=100034πR3

therefore, r=10R

Now, the surface area of small drop =4πr2=4π(R/10)2

=4×3.14×(1×10