Question

Work done on a liquid drop to increase its radius from $3\text{cm}$ to $5\text{cm}$ is $\pi \times {10}^{-4}\text{J}$. Find the surface tension of liquid.

• A. $\frac{1}{8}\text{N/m}$
• B. $\frac{1}{4}\text{N/m}$
• C. $\frac{1}{64}\text{N/m}$
• D. $\frac{1}{32}\text{N/m}$

Answer 1

The correct option is C $\frac{1}{64}\text{N/m}$

As we know,
Liquid drop has only one layer.
$\therefore$ Work done in formation of liquid drop;
$W=4\pi T(r_f^2-r_i^2)$
where,
Surface tension of liquid is $T$
Final radius of liquid drop,$r_f=5\times {10}^{-2}\text{m}$
Initial radius of liquid drop,$r_i=3\times {10}^{-2}\text{m}$

$\Rightarrow 4\pi \times T\times ({0.05}^2-{0.03}^2)=\pi \times {10}^{-4}\text{J}$
$\Rightarrow T=\frac{1}{64}\text{N/m}$