Two mercury drops,each with same radius r,mer | vedclass

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

Question

(D) Let the radius of the bigger drop be $R$. Since the volume remains constant during the merger of two drops,we have:$2 	imes \frac{4}{3} \pi r^3 =

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the radius of the bigger drop be $R$. Since the volume remains constant during the merger of two drops,we have:$2 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$The surface energy $E$ of the bigger drop is given by the product of surface tension $T$ and its surface area $A = 4 \pi R^2$:$E = T 	imes 4 \pi R^2$Substituting $R = 2^{1/3} r$ into the equation:$E = T 	imes 4 \pi (2^{1/3} r)^2$$E = T 	imes 4 \pi 	imes 2^{2/3} r^2$$E = 4 	imes 2^{2/3} \pi r^2 T$Since $4 = 2^2$,we have $2^2 	imes 2^{2/3} = 2^{2 + 2/3} = 2^{8/3}$.Therefore,$E = 2^{8/3} \pi r^2 T$.

Two mercury drops,each with same radius $r$,merge to form a bigger drop. If $T$ is the surface tension of mercury,then the surface energy of the bigger drop is given by

Similar Questions

$A$ big drop is formed by coalescing $1000$ small droplets of water. The ratio of surface energy of $1000$ droplets to that of the energy of the big drop is $\frac{10}{x}$. The value of $x$ is . . . . . . .

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

The energy needed for breaking a liquid drop of radius '$R$' into '$n$' droplets each of radius '$r$' is [$T$ = surface tension of the liquid]

Two small droplets combine to form a single large drop. What is the ratio of the surface energy of the small droplets to the surface energy of the large drop?

Work done in increasing the size of a soap bubble from a radius of $3\, cm$ to $5\, cm$ is (Surface tension of soap solution $= 0.03\, Nm^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module