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) Let $T$ be the surface tension of mercury.Total surface energy before coalescence $(E_1)$: Since there are two drops of radius $R$,$E_1 = 2 	imes

Vedclass · Accepted Answer

$2^{1/3}: 1$

Vedclass · Answer

(C) Let $T$ be the surface tension of mercury.Total surface energy before coalescence $(E_1)$: Since there are two drops of radius $R$,$E_1 = 2 	imes (4 \pi R^2 T) = 8 \pi R^2 T$.Let the radius of the new large drop be $R'$. By conservation of volume: $\frac{4}{3} \pi R'^3 = 2 	imes \frac{4}{3} \pi R^3$,which gives $R'^3 = 2R^3$,or $R' = 2^{1/3} R$.Total surface energy after coalescence $(E_2)$: $E_2 = 4 \pi R'^2 T = 4 \pi (2^{1/3} R)^2 T = 4 \pi (2^{2/3}) R^2 T$.The ratio of surface energies is $\frac{E_1}{E_2} = \frac{8 \pi R^2 T}{4 \pi (2^{2/3}) R^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.Thus,the ratio is $2^{1/3} : 1$.

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

$A$ circular conducting loop of radius $R$ carries a current $I.$ Another straight infinite conductor carrying current $I$ passes through the diameter of this loop as shown in the figure. The magnitude of force exerted by the straight conductor on the loop is

The maximum value of the function $f(x) = x^3 - 12x^2 + 36x + 17$ in the interval $[1, 10]$ is:

Data Entry - Physics

From which plant did $D.J. Ivanowsky$ discover viruses?

The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is

Vedclass Test Series

Exam Paper Generator

Online Exam Module