The surface tension of liquid is 0.5 N/m. If a film is held on a ring of area 0.02 ${\mathrm{m}}^2$, its surface energy is

(a) $5\times {10}^{-2} \mathrm{joule}$                         (b) $2\times {10}^{-2} \mathrm{joule}$

(c) $4\times {10}^{-4} \mathrm{joule}$                         (d) $0.8\times {10}^{-1} \mathrm{joule}$

A liquid drop of diameter D breaks upto into 27 small drops of equal size. If the surface tension of the liquid is $\mathrm{\sigma }$, then change in surface energy is 

(a) ${\mathrm{\pi D}}^2\mathrm{\sigma }$                          (b) $2{\mathrm{\pi D}}^2\mathrm{\sigma }$

(c) $3{\mathrm{\pi D}}^2\mathrm{\sigma }$                         (d) $4{\mathrm{\pi D}}^2\mathrm{\sigma }$

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is 

(1) 1000 : 1               

(2) 1 : 1000

(3) 10 : 1                   

(4) 1 : 10

A big drop of radius R is formed by 1000 small droplets of water, then the radius of small drop is

(1) R/2                                     

(2) R/5

(3) R/6                                     

(4) R/10

8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is 
(1) 1 : 10                     

(2) 1 : 15

(3) 1 : 20                     

(4) 1 : 25

When two small bubbles join to form a bigger one, energy is

(1) Released

(2) Absorbed

(3) Both (1) and (2)

(4) None of these

A water film is formed between the two straight parallel wires, each of length \(10~\text{cm}\), kept at a separation of \(0.5~\text{cm}.\)Now, the separation between them is increased by \(1~\text{mm}\) without breaking the water film. The work done for this is
(surface tension of water =\(7.2\times 10^{-2}~\text{N/m}\))
1. \(7.22\times 10^{-6}~\text{J}\)
2. \(1.44\times 10^{-5}~\text{J}\)
3. \(2.88\times 10^{-5}~\text{J}\)
4. \(5.76\times 10^{-5}~\text{J}\)

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is $0.465 \mathrm{J}/{\mathrm{m}}^2$) 

(1)$23.4 \mathrm{\mu J}$

(2) $18.5 \mathrm{\mu J}$

(3) $26.8 \mathrm{\mu J}$

(4) $16.8 \mathrm{\mu J}$

If two soap bubbles of equal radii r coalesce then the radius of curvature of interface between two bubbles will be

(1) r                                 

(2) 0

(3) Infinity                         

(4) 1/2r

The angle of contact between glass and mercury is

(1) 0°

(2) 30°

(3) 90°

(4) 135°