Question

The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is

• A. $ 1:16 $
• B. $ 8:1 $
• C. $ 1:4 $
• D. $ 1:64 $

Answer 1

The Correct Option is D

Excess pressure inside a spherical drop of water $ P=\frac{2T}{R} $ Given: $ {{p}_{1}}=4{{p}_{2}} $ $ \frac{2T}{{{R}_{1}}}=4\times \frac{2T}{{{R}_{2}}} $ Or $ {{R}_{2}}=4{{R}_{1}} $ Now, $ \frac{{{m}_{1}}}{{{m}_{2}}}=\frac{4\pi R_{1}^{3}{{d}_{1}}}{4\pi R_{2}^{3}{{d}_{2}}} $ Or $ \frac{{{m}_{1}}}{{{m}_{2}}}=\frac{R_{1}^{3}}{R_{2}^{3}} $ Or $ \frac{{{m}_{1}}}{{{m}_{2}}}=\frac{1}{64} $