A big drop is formed by coalescing 1000 small | vedclass

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

Question

(D) Let the radius of each small droplet be $r$ and the radius of the big drop be $R$.Since the volume remains conserved,the volume of the big drop eq

Vedclass · Accepted Answer

$\frac{1}{10}$ th

Vedclass · Answer

(D) Let the radius of each small droplet be $r$ and the radius of the big drop be $R$.Since the volume remains conserved,the volume of the big drop equals the sum of the volumes of $1000$ small droplets:$\frac{4}{3} \pi R^3 = 1000 	imes \frac{4}{3} \pi r^3$$R^3 = 1000 r^3$$R = 10r$The initial surface energy $U_i$ of $1000$ small droplets is:$U_i = 1000 	imes (4 \pi r^2 S)$,where $S$ is the surface tension.The final surface energy $U_f$ of the big drop is:$U_f = 4 \pi R^2 S$Substituting $R = 10r$:$U_f = 4 \pi (10r)^2 S = 100 	imes (4 \pi r^2 S)$Comparing $U_f$ and $U_i$:$U_f = \frac{100 	imes (4 \pi r^2 S)}{1000 	imes (4 \pi r^2 S)} U_i$$U_f = \frac{1}{10} U_i$Thus,the surface energy becomes $\frac{1}{10}$ th of the initial value.

$A$ big drop is formed by coalescing $1000$ small droplets of water. The surface energy will become :

Similar Questions

$A$ bubble of $8$ mm diameter is formed in the air. The surface tension of soap solution is $30$ dynes/cm. The excess pressure inside the bubble is ........ $dynes/cm^2$.

The excess of pressure,due to surface tension,on a spherical liquid drop of radius $R$ is proportional to

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

Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).

What happens when a soap bubble is charged?

Vedclass Test Series

Exam Paper Generator

Online Exam Module