Question

Two mercury droplets of radii $0.1cm$ and $0.2cm$ collapse into one single drop. What amount of energy is released/ absorbed? The surface tension of mercury $T=435.5\times {10}^{-3}N{m}^{-1}$.

• A. $32.23\times {10}^{-7}$ (Released)
• B. $32.23\times {10}^{-7}$ (Absorbed)
• C. $64.46\times {10}^{-7}$ (Absorbed)
• D. $64.46\times {10}^{-7}$ (Released)

Answer 1

Answer: (A)

$32.23\times {10}^{-7}$ (Released)

Let $R$ be the radius of big single drop formed when two small drops of radii $r_1$ and $r_2$ collapse together.

Then volume of bid drop $=$ volume of two small drops

$\therefore \frac{4}{3}\pi R^1=\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^3$

$R^3=r_1^3+r_2^3=(0.1{)}^3+(0.2{)}^3=0.001+0.008=0.009{cm}^3$

$\therefore R=0.21cm$

Change in surface area $=4\pi R^2-(4\pi r_1^2+4{\pi }_2^2)$

Energy relased $=ST\times$ change in surface area $=T\times [4\pi R^2-(4\pi r_1^2+4\pi r_2^2)]=4\pi T[R^2-(r_1^2+r_2^2)]$

$4\times 3.142\times 435.5\times {10}^{-3}\left[\right(0.21{)}^2-\left\{\right(0.1{)}^2+\left(0.2{)}^2\right\}]=-32.33\times {10}^{-7}J$

$-$ sign shows that energy is released , Hence option A is correct