The excess pressure inside a spherical drop o | vedclass

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

Question

(D) The excess pressure inside a spherical drop of radius $r$ is given by $P = \frac{2T}{r}$,where $T$ is the surface tension.For drop $A$,$P_A = \fra

Vedclass · Accepted Answer

$1: 64$

Vedclass · Answer

(D) The excess pressure inside a spherical drop of radius $r$ is given by $P = \frac{2T}{r}$,where $T$ is the surface tension.For drop $A$,$P_A = \frac{2T}{r_A}$.For drop $B$,$P_B = \frac{2T}{r_B}$.Given that $P_A = 4P_B$,we have $\frac{2T}{r_A} = 4 \left( \frac{2T}{r_B} \right)$.This simplifies to $\frac{1}{r_A} = \frac{4}{r_B}$,or $\frac{r_A}{r_B} = \frac{1}{4}$.The mass of a drop is given by $m = V \rho = \left( \frac{4}{3} \pi r^3 \right) \rho$.Since both are water drops,the density $\rho$ is the same.Therefore,the ratio of masses is $\frac{m_A}{m_B} = \frac{r_A^3}{r_B^3} = \left( \frac{r_A}{r_B} \right)^3$.Substituting the ratio of radii,$\frac{m_A}{m_B} = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.

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

Similar Questions

Let $R_{1}$ and $R_{2}$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

Write the equation of excess pressure for a liquid drop.

When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is:

Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

Let $n$ be the number of liquid drops,each with surface energy $E$. These drops join to form a single drop. In this process:

Vedclass Test Series

Exam Paper Generator

Online Exam Module