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

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) ${\mathrm{A}}^2{\mathrm{T}}^2$