Question

8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of:

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

The correct option is C 4

Surface energy = surface tension x area of surface
For 1 drop, volume = $\frac{4}{3}\pi R^3$ if R = radius of drop.
Total volume of 8 drops = 8. $\frac{4}{3}\pi R^3$ = $\frac{4}{3}\pi (2R{)}^3$
R' = 2R, new radius of big drop
New area = 4$\pi 4R^2$ = 4 $\times$ old area
Energy $\propto$ area
$E_1$ = 4$\pi R^2$.....................(1)
$E_2$ = 4.4$\pi R^2$.....................(2)
From equation (1) and (2) we get, $E_2$ = $4E_1$