The property utilized in the manufacture of lead shots is

1. Specific weight of liquid lead

2. Specific gravity of liquid lead

3. Compressibility of liquid lead

4. Surface tension of liquid lead

The dimensions of surface tension are

1. $\left[{\mathrm{MLT}}^{-1}\right]$                 

2. $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$

3. $\left[{\mathrm{ML}}^0{\mathrm{T}}^{-2}\right]$                 

4. $\left[{\mathrm{ML}}^1{\mathrm{T}}^{-2}\right]$This quantity S is the magnitude of surface
tension

A wooden stick 2 m long is floating on the surface of the water. The surface tension of water is 0.07 N/m. By putting soap solution on one side of the stick, the surface tension is reduced to 0.06 N/m. The net force on the stick due to surface tension will be:

Surface tension may be defined as 

1. The work done per unit area in increasing the surface area of a liquid under isothermal condition

2. The work done per unit area in increasing the surface area of a liquid under adiabatic condition

3. The work done per unit area in increasing the surface area of a liquid under both isothermal and adiabatic conditions

4. Free surface energy per unit volume

The energy needed to break a drop of radius \(R\) into \(n\) drops of radii \(r\) is given by:
1. \(4 πT ( nr ^2 - R ^2 )\)
2. \(\frac{4}{3} \pi \left(r^{3} n - R^{2}\right)\)
3. \(4 πT \left(R^{2} -nr^{2}\right)\)
4. \(4 πT \left(nr^{2}+R^{2} \right)\)$$

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

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

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

1. $9.58\times {10}^{-5} \mathrm{J}$                 

2. $8.95\times {10}^{-5} \mathrm{J}$

3. $5.89\times {10}^{-5} \mathrm{J}$                   

4. $5.98\times {10}^{-5} \mathrm{J}$

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

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