Gene Kelly physics

We have for any spherical drop the excess pressure in it is given by

P= $\frac {2T}{R}$ (Lets say eqn 1)

Where T is Surface Tension and R is the radius of the spherical drop

Also we have volume of the drop = $\frac{4}{3}$ $\pi$ $R^3$ = $4.2 \times10^{-9}$ $m^3$

This gives us on solving R= $10^{-3}$ meters

Hence Putting the value of R in eqn 1 we get T= 0.14 $\frac{N}{m}$

pressure difference(P)=2 surface tension(T)/radius of drop(r).....(i) from V=4/3{pi} r^3 r=10^(-3) Also P=280 Pa Thus putting in eqn. (i) T=0.14 N/m

Excess pressure inside a liquid drop is related to surface tension by the formula: $\Delta p=\frac{2T}{R}$ where T is the surface tension and R is the radius of drop.

The radius can be calculated from the volume given in the question, i.e $\displaystyle r=\sqrt[3] {\frac{3 \times 4.2 \times 10^{-9}}{4\pi}}m$

Plugging in the values of $R$ and $\Delta p$ in the above formula and solving for T, $\fbox{T=0.14012 N/m}$

The two forces that act in a half-hemisphere are the one caused by the difference of pression between the inside and the outside and the one caused by the surface tension. The second one acts on the line that delimits the half-hemisphere and it is equal to $2 \pi R\times T$ , being $T$ the surface tension. The first acts along all the half-hemisphere surface, but it balances the second only in one direction. Therefore, we must considere it only in the direction of the second one. The force caused by the pression difference is equal to $\pi R^2 \times \Delta P$ , being $\Delta P$ the difference of pression. From the equation $V=\frac{4 \pi R^3}{3}$ we discover that $R=10^{-3} m$ . Writing that the first force is equal to the second one, we discover that: $T=\frac{P \times R}{2}=0.140 N/m$ .

as the excess pressure inside drop is 2T/r where t is surface tension and r is radius drop. here r is aprox 1mm so on equating T is 0.14.

Net pressure force equals to difference between internal pressure force and external pressure force,

Let internal pressure denoted by lowercase letter p and external pressure by P.

The area on which the pressure forces exert is given by the projected area of half the spherical area, $\pi R^2$ ,

Surface tension force of a sphere is given by the product of surface tension and the circumference, $2\pi RT$ .

Thus we can equate:

$p\pi R^2-P\pi R^2=dP\pi R^2= 2\pi RT$ .

Hence $T= dP R/2$

Since $V=(4/3)\pi R^3$

we get R= 1.00E-3

Substitute the value of R back into the equation, T=0.14N/m.

The surface tension $\sigma$ holding the raindrop together exactly balances the pressure force trying to blow the raindrop apart. Since the forces exactly balance, the raindrop is at a minimum in energy, i.e. the work needed to change the radius slightly is zero. If we change the radius by some small $dr$ , the work is $dW=\sigma \,dA-\Delta p\, dV=\sigma 8 \pi r \,dr-4\pi r^2 \Delta p \,dr=0$ . We therefore can solve for $\Delta p=2 \sigma/r$ . The radius of our raindrop can be calculated from the volume, $r=0.001~\mbox{m}$ , and so we can determine that $\sigma=0.14~\mbox{N/m}$ .