Two mercury drops (each of radius r) merge to | vedclass

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

Question

(D) When two drops of radius $r$ merge to form a bigger drop of radius $R$, the volume remains conserved.
Volume of two small drops = Volume of one b

Vedclass · Accepted Answer

$2^{8/3} \pi r^2 T$

Vedclass · Answer

(D) When two drops of radius $r$ merge to form a bigger drop of radius $R$, the volume remains conserved.
Volume of two small drops = Volume of one big drop
$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 2r^3$
$R = 2^{1/3} r$
The surface energy $E$ of a drop is given by $E = 	ext{Surface Area} 	imes 	ext{Surface Tension} = 4 \pi R^2 T$.
Substituting the value of $R$:
$E = 4 \pi (2^{1/3} r)^2 T$
$E = 4 \pi (2^{2/3} r^2) T$
$E = 2^2 	imes 2^{2/3} \pi r^2 T$
$E = 2^{2 + 2/3} \pi r^2 T$
$E = 2^{8/3} \pi r^2 T$.

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

Similar Questions

$A$ thin film of water is formed between two straight parallel wires each of length $8 \ cm$ separated by a distance of $0.6 \ cm$. The work done to increase the distance between the wires to $0.8 \ cm$ is (Surface tension of water $= 0.07 \ N/m$) (in $\mu J$)

Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

What is surface energy?

$A$ soap bubble of radius $\frac{1}{\sqrt{\pi}} \text{ cm}$ is expanded to radius $\frac{3}{\sqrt{\pi}} \text{ cm}$. The surface tension of the soap solution is $25 \text{ dyne/cm}$. The work done during expansion in $\text{erg}$ is:

Surface tension of a soap bubble is $2.0 \times 10^{-2} \; N m^{-1}$. Work done to increase the radius of the soap bubble from $3.5 \; cm$ to $7 \; cm$ will be $......... \times 10^{-4} \; J$ [Take $\pi = \frac{22}{7}$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module