If 1000 droplets of water of surface tension 0 . 07N m ^ - 1 ....

Let the radius of bigger droplet is $R$, then by volume conservation:

\left(\dfrac{4}{3}\pi{R}^{3}\right) = 1000\left(\dfrac{4}{3}\pi{\left(1\right)}^{3}\right)$</p><p>$⇒ R = 10mm$</p><p>Final potential energy:</p><p>${U}_{f} = 1000\left(4\pi{R}^{2}T\right) = 1000 \times 4\pi{\left(10 \times {10}^{- 3}\right)}^{2}T$</p><p>Initial potential energy:</p><p>${U}_{i} = 4\pi \times {\left(r\right)}^{2}T = 4\pi \times {\left({10}^{- 3}\right)}^{2}T$</p><p>Therefore, change in potential energy,</p><p>$∆ E = T \times 1000 \times 4\pi{\left({10}^{- 3}\right)}^{2} - T \times 4\pi{\left(10 \times {10}^{- 3}\right)}^{2}$</p><p>$∆ E = 4 \times \pi \times 7 \times {10}^{- 2}\left[1000 - 100\right] \times {10}^{- 6}$</p><p>$∆ E = 7 . 92 \times {10}^{- 4}J$