If 1000 droplets of water of surface tension 0.07 N m$$^{-1}$$, having same radius 1 mm each, combine to form a single drop. In the process the released surface energy is- (Take $$\pi = \dfrac{22}{7}$$)

Given: 1000 droplets each of radius $$r = 1$$ mm $$= 10^{-3}$$ m, surface tension $$T = 0.07$$ N/m.

To begin, finding the radius of the combined drop,

By conservation of volume:

$$ 1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 $$

$$ R = 10r = 10 \times 10^{-3} = 10^{-2} \text{ m} $$

Next, change in surface area,

Initial total surface area = $$1000 \times 4\pi r^2 = 4000\pi \times 10^{-6}$$ m$$^2$$

Final surface area = $$4\pi R^2 = 4\pi \times 10^{-4}$$ m$$^2$$

$$ \Delta A = 4000\pi \times 10^{-6} - 4\pi \times 10^{-4} = 4\pi(1000 \times 10^{-6} - 10^{-4}) $$

$$ = 4\pi(10^{-3} - 10^{-4}) = 4\pi \times 9 \times 10^{-4} = 36\pi \times 10^{-4} $$

From this, released surface energy,

$$ E = T \times \Delta A = 0.07 \times 36 \times \frac{22}{7} \times 10^{-4} $$

$$ = 0.01 \times 36 \times 22 \times 10^{-4} = 0.01 \times 792 \times 10^{-4} = 7.92 \times 10^{-4} \text{ J} $$