What is the ratio of the surface energy of 1 | vedclass

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

Question

(D) The volume of the liquid remains constant,meaning the volume of $1000$ small drops is equal to the volume of one large drop.$n \cdot \frac{4}{3} \

Vedclass · Accepted Answer

$1:100$

Vedclass · Answer

(D) The volume of the liquid remains constant,meaning the volume of $1000$ small drops is equal to the volume of one large drop.$n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$1000 r^3 = R^3 \implies R = 10r$Therefore,the ratio of radii is $\frac{r}{R} = \frac{1}{10}$.The surface energy $E$ is given by $E = T \cdot A$,where $T$ is surface tension and $A$ is the surface area.$\frac{E_{	ext{small}}}{E_{	ext{large}}} = \frac{4 \pi r^2 T}{4 \pi R^2 T} = \left( \frac{r}{R} ight)^2$Substituting the ratio: $\left( \frac{1}{10} ight)^2 = \frac{1}{100}$.Thus,the ratio is $1:100$.

What is the ratio of the surface energy of $1$ small drop to $1$ large drop,if $1000$ small drops combine to form $1$ large drop?

Similar Questions

$A$ liquid drop at temperature $t$,isolated from its surroundings,breaks into a number of droplets. The temperature of the droplets will be

Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

The surface tension of a liquid is $5 \,N/m$. If a film is held on a ring of area $0.02 \,m^2$,its surface energy is about ........... $J$.

$A$ small liquid drop of radius $R$ is divided into $27$ identical liquid drops. If the surface tension is $T$,then the work done in the process will be

$n$ drops of a liquid,each with surface energy $E$,join to form a single drop. Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module