Question

The work done to get 'n' smaller equal size spherical drops from a bigger size spherical drop of water is proportional to –

• A. $\frac{1}{n^{2/3}}$ –1
• B. $\frac{1}{n^{1/3}}$ –1
• C. n^1/3 –1
• D. n^4/3 – 1

Answer 1

Answer: (C)

n^1/3 –1

Answer 2

$\frac{4}{3}$pR³ = n ×$\frac{4}{3}$

pr³ or r = $\frac{R}{n^{1/3}}$

Now work done = increase in area × surface tension

= (4pr²n – 4pR²) × T = $4\pi\left( \frac{R^{2}}{n^{2/3}}n - R^{2} \right) \times T$

µ (n^1/3 – 1)