A liquid drop of radius R is broken into n id | vedclass

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

Question

(B) The volume of the original drop is equal to the sum of the volumes of the $n$ smaller droplets.Let $r$ be the radius of each small droplet.$n \cdo

Vedclass · Accepted Answer

$4 \pi R^2(n^{1/3}-1) T$

Vedclass · Answer

(B) The volume of the original drop is equal to the sum of the volumes of the $n$ smaller droplets.Let $r$ be the radius of each small droplet.$n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$r^3 = \frac{R^3}{n} \implies r = \frac{R}{n^{1/3}}$Work done $W$ is equal to the change in surface area multiplied by the surface tension $T$.$W = (A_{	ext{final}} - A_{	ext{initial}}) T$$W = (n \cdot 4 \pi r^2 - 4 \pi R^2) T$Substitute $r = R \cdot n^{-1/3}$:$W = (n \cdot 4 \pi (R \cdot n^{-1/3})^2 - 4 \pi R^2) T$$W = (4 \pi R^2 \cdot n \cdot n^{-2/3} - 4 \pi R^2) T$$W = 4 \pi R^2 (n^{1/3} - 1) T$

$A$ liquid drop of radius $R$ is broken into $n$ identical small droplets. The work done is $[T = \text{surface tension of the liquid}]$

Similar Questions

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times the volume of the second,where $n$ is:

$1000$ small water drops,each of radius $r$ and charge $q$,coalesce together to form one large spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of:

Two small drops of mercury each of radius $r$ coalesce to form a single large drop of radius $R$. The ratio of the total surface energies before and after the change is

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3}\, N/m$,the difference of pressure between the inside and outside of the spherical drop is ........ $N/m^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module