Question

An elliptically shaped ring of dimensions shown in figure just touches, the horizontal surface of a liquid of surface tension S. The force required to pull the ring away from the liquid surface is

• A. $2\pi (\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S$
• B. $\pi ({{a}_{1}}+{{b}_{1}}+{{a}_{2}}+{{b}_{2}})S$
• C. $\pi \left( \frac{{{a}_{1}}+{{a}_{2}}}{2}+\frac{{{b}_{1}}+{{b}_{2}}}{2} \right)S$
• D. $\sqrt{2\pi }(\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S$

Answer 1

Answer: (A)

$2\pi (\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S$

Answer 2

Internal mean radius ${{r}_{1}}\sqrt{{{a}_{1}}{{b}_{1}}}$ Internal circumference of the ring $=2\pi {{r}_{1}}=2\pi \sqrt{{{a}_{1}}{{b}_{1}}}$ External mean radius ${{r}_{2}}=\sqrt{{{a}_{2}}{{b}_{2}}}$ External circumference of the ring $=2\pi {{r}_{2}}=2\pi \sqrt{{{a}_{2}}{{b}_{2}}}$ Thus, force required $F=2\pi \sqrt{{{a}_{1}}{{b}_{1}}S}+2\pi \sqrt{{{a}_{2}}{{b}_{2}}}\,S$ $=2\pi (\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S$