In air,a charged soap bubble of radius R brea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(NONE) Let the charge on the big bubble be $Q$ and its radius be $R$. The charge density is $\sigma = \frac{Q}{4\pi R^2}$.
The mechanical force per u

Vedclass · Accepted Answer

$16: 1$

Vedclass · Answer

(NONE) Let the charge on the big bubble be $Q$ and its radius be $R$. The charge density is $\sigma = \frac{Q}{4\pi R^2}$.
The mechanical force per unit area (electrostatic pressure) on a charged bubble is given by $P = \frac{\sigma^2}{2\epsilon_0} = \frac{Q^2}{2\epsilon_0 (4\pi R^2)^2} = \frac{Q^2}{32\pi^2 \epsilon_0 R^4}$.
When the bubble breaks into $64$ small bubbles,the volume is conserved: $\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$,which gives $R^3 = 64r^3$,so $R = 4r$.
The total charge $Q$ is conserved,so each small bubble has charge $q = \frac{Q}{64}$.
The electrostatic pressure on a small bubble is $p = \frac{q^2}{32\pi^2 \epsilon_0 r^4} = \frac{(Q/64)^2}{32\pi^2 \epsilon_0 (R/4)^4} = \frac{Q^2 / 64^2}{32\pi^2 \epsilon_0 (R^4 / 256)} = \frac{Q^2}{32\pi^2 \epsilon_0 R^4} 	imes \frac{256}{4096} = P 	imes \frac{1}{16}$.
Therefore,the ratio of the mechanical force per unit area of the big bubble to that of a small bubble is $P/p = 16:1$.

In air,a charged soap bubble of radius $R$ breaks into $64$ small soap bubbles of equal radius $r$. The ratio of mechanical force per unit area of the big soap bubble to that of a small bubble is

Similar Questions

If $W$ amount of work is required to form a bubble of volume $V$,then how much work is required to form a bubble of volume $2V$?

The excess pressure inside a spherical water drop $A$ is four times that of another water drop $B$. Then, the ratio of the mass of water drop $A$ to that of drop $B$ is

The work done to break a spherical drop of radius $R$ into $n$ drops of equal size is proportional to .............

$A$ big drop of radius $R$ is formed by $1000$ small droplets of water. The radius of each small droplet is:

The excess pressure due to surface tension in a spherical liquid drop of radius $r$ is directly proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module