Question 30: Properties of Matter

Question

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density $1000 \mathrm{~kg} / \mathrm{m}^3$. If the pressure inside the bubble is $2100 \mathrm{~N} / \mathrm{m}^2$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

$$ P_{	ext{bubble}} = P_{	ext{atm}} + ho g h + \frac{2S}{r} $$
Given that the pressure inside the bubble is $$P_{	ext{atm}} + 2100\,	ext{N/m}^2$$ and the liquid pressure at a depth of 20 cm is
$$ P_{	ext{liquid}} = P_{	ext{atm}} + ho g h = P_{	ext{atm}} + 1000 	imes 10 	imes 0.20 = P_{	ext{atm}} + 2000\,	ext{N/m}^2, $$
the pressure difference due solely to surface tension is
$$ \Delta P = \Bigl(P_{	ext{bombubble}} - P_{	ext{liquid}}\Bigr) = \bigl(P_{	ext{atm}}+2100\bigr) - \bigl(P_{	ext{atm}}+2000\bigr) = 100\,	ext{N/m}^2. $$
The Laplace pressure difference for a bubble (which has one interface) is given by
$$ \Delta P = \frac{2S}{r}. $$
The given bubble radius is 0.1 cm, which in SI units is
$$ r = 0.1\,	ext{cm} = 0.001\,	ext{m}. $$
Substituting the known values into the Laplace equation gives
$$ \frac{2S}{0.001} = 100. $$
Solving for $$S$$:
$$ 2S = 100 	imes 0.001 = 0.1, $$
$$ S = \frac{0.1}{2} = 0.05\,	ext{N/m}. $$
Thus, the surface tension of the liquid is
$$ \boxed{0.05\,	ext{N/m}}. $$

Answer

0.02

Answer

0.05

Answer

0.25

Answer

0.1

Properties of Matter

Choose the correct answer: