The surface tension of a liquid at its boiling point

(1) Becomes zero

(2) Becomes infinity

(3) is equal to the value at room temperature

(4) is half to the value at the room temperature

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) $16.8 \mathrm{\mu J}$

Two small drops of mercury, each of radius r, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

1. $1:2^{1/3}$

2. $2^{1/3} :1$

3. 2:1

4. 1:2