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

1. $5\times {10}^{-2}<mspace\; width="1em"></mspace>\mathrm{joule}$                         
2. $2\times {10}^{-2}<mspace\; width="1em"></mspace>\mathrm{joule}$
3. $4\times {10}^{-4}<mspace\; width="1em"></mspace>\mathrm{joule}$                       
4. $0.8\times {10}^{-1}<mspace\; width="1em"></mspace>\mathrm{joule}$

If the surface tension of a liquid is T, the gain in surface energy for an increase in liquid surface by A is

1. ${\mathrm{AT}}^{-1}$                       

2. $\mathrm{AT}$

3. ${\mathrm{A}}^2\mathrm{T}$                         

4. ${\mathrm{A}}^2{\mathrm{T}}^2$

Two droplets merge with each other and forms a large droplet. In this process

1. Energy is liberated

2. Energy is absorbed

3. Neither liberated nor absorbed

4. Some mass is converted into energy

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

Work done in splitting a drop of water of 1 mm radius into ${10}^6$ droplets is (Surface tension of water $=72\times {10}^{-3}<mspace\; width="1em"></mspace>\mathrm{J}/{\mathrm{m}}^2$)

1. $9.58\times {10}^{-5}<mspace\; width="1em"></mspace>\mathrm{J}$                 

2. $8.95\times {10}^{-5}<mspace\; width="1em"></mspace>\mathrm{J}$

3. $5.89\times {10}^{-5}<mspace\; width="1em"></mspace>\mathrm{J}$                   

4. $5.98\times {10}^{-5}<mspace\; width="1em"></mspace>\mathrm{J}$

A spherical drop of oil of radius 1 cm is broken into 1000 droplets of equal radii. If the surface tension of oil is 50 dynes/cm, the work done is 

1. 18 $\pi$ ergs                     

2. 180 $\pi$ ergs

3. 1800 $\pi$ ergs                 

4. 8000 $\pi$ ergs

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) $4{\mathrm{\pi D}}^2\mathrm{\sigma }$

The potential energy of a molecule on the surface of liquid compared to one inside the liquid is  -
1. Zero                                             

2. Smaller

3. The same                                     

4. Greater

The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T= surface tension of the solution)

1. $4\mathrm{\pi }({\mathrm{D}}^2-{\mathrm{d}}^2)\mathrm{T}$                     

2. $8\mathrm{\pi }({\mathrm{D}}^2-{\mathrm{d}}^2)\mathrm{T}$

3. $\mathrm{\pi }({\mathrm{D}}^2-{\mathrm{d}}^2)\mathrm{T}$                       

4. $2\mathrm{\pi }({\mathrm{D}}^2-{\mathrm{d}}^2)\mathrm{T}$