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

Answer: (D)

$1:64$

Answer 2

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 $\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{1}{64}$