A spherical drop of water has a radius of 1 mm. If the surface tension of water is $70\times {10}^{-3}$ N/m, the difference in pressures inside and outside the spherical drop is:

The property of surface tension of the liquid is due to:

Small droplets of liquid are usually more spherical in shape than the larger drops of the same liquid because:

If the excess pressure inside a soap bubble is balanced by an oil column of height of \(2~\text{mm},\) then the surface tension of the soap solution will be:
(the radius of the soap bubble, \(r=1~\text{cm}\) and density of oil, \(d=0.8~\text{gm/cm}^3\) )
1. \(3.9~\text {N/m}\) 
2. $\mathrm{}$\(3.9\times 10^{-2}~\text{N/m}\)$\mathrm{}$
3. $\mathrm{}$\(3.9\times 10^{-3}~\text{N/m}\)$\mathrm{}$
4. \(3.9​​​​~\text{dyne/m}\) 

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. \(4 πT \left(nr^{2}+R^{2} \right)\)$$

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:

The surface tension of a liquid is \(5\) N m^-1. If a film is held on a ring of area \(0.02\) ${\mathrm{m}}^2$, then its total surface energy is:
1. \(3\times\) ${10}^{-2}$ J
2. \(2\times\) ${10}^{-1}$ J
3. \(3\times\) ${10}^{-3}$J
4. ${10}^{-1}$ J

Two soap bubbles are connected by a tube as shown in the figure. Which of the following statement is true?
                                  

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then:\(\text { Energy }=4 V T\left(\frac{1}{r}-\frac{1}{R}\right) \text { is released } \)