Two small drops of mercury,each of radius R,c | vedclass

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

Question

(C) Surface energy $E$ is proportional to the surface area $A$,where $E = T 	imes A$ ($T$ is surface tension).Initial surface area of two drops: $A_{

Vedclass · Accepted Answer

$2^{1/3} : 1$

Vedclass · Answer

(C) Surface energy $E$ is proportional to the surface area $A$,where $E = T 	imes A$ ($T$ is surface tension).Initial surface area of two drops: $A_{	ext{before}} = 2 	imes (4 \pi R^2) = 8 \pi R^2$.When they coalesce into a single large drop of radius $R_0$,the volume remains conserved:$2 	imes (\frac{4}{3} \pi R^3) = \frac{4}{3} \pi R_0^3 \Rightarrow R_0^3 = 2R^3 \Rightarrow R_0 = R(2)^{1/3}$.Final surface area: $A_{	ext{after}} = 4 \pi R_0^2 = 4 \pi (R \cdot 2^{1/3})^2 = 4 \pi R^2 \cdot 2^{2/3}$.The ratio of surface energies is the ratio of surface areas:$\frac{E_{	ext{before}}}{E_{	ext{after}}} = \frac{8 \pi R^2}{4 \pi R^2 \cdot 2^{2/3}} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3} : 1$.

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

Similar Questions

What are the products formed when chlorine is passed through an aqueous hypo solution?

The height at which the weight of a body becomes $1/16^{th}$ of its weight on the surface of the earth (radius $R$) is: (in $R$)

In $chloroethane$,the carbon bearing halogen is bonded to $...$ hydrogen $(s)$. It is called $......$ alkyl halide.

$H_{3}PO_{3}$ can be represented by structures $1$ and $2$ shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing $H_{3}PO_{3}$? If not,give reasons for the same.

Vedclass Test Series

Exam Paper Generator

Online Exam Module