An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density $$10^3$$ kg/m$$^3$$. The pressure inside the bubble is 1100 Nm$$^{-2}$$ greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? ($$g = 9.8$$ ms$$^{-2}$$)

We are given:

• Radius of air bubble, $$ r = 0.1 \, \text{cm} $$. Convert to meters: $$ 0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 0.001 \, \text{m} $$.
• Surface tension, $$ S = 0.06 \, \text{N/m} $$.
• Density of liquid, $$ \rho = 10^3 \, \text{kg/m}^3 = 1000 \, \text{kg/m}^3 $$.
• Excess pressure inside bubble over atmospheric pressure, $$ \Delta P = 1100 \, \text{N/m}^2 $$.
• Acceleration due to gravity, $$ g = 9.8 \, \text{m/s}^2 $$.

We need to find the depth $$ h $$ (in meters) of the bubble below the surface of the liquid.

For an air bubble inside a liquid, the excess pressure inside the bubble over the atmospheric pressure is the sum of two components:

1. The pressure due to the depth of the bubble: $$ \rho g h $$.
2. The pressure due to surface tension: $$ \frac{2S}{r} $$.

Therefore, the equation for excess pressure is:

$$ \Delta P = \rho g h + \frac{2S}{r} $$

Substitute the given values:

$$ 1100 = (1000) \times (9.8) \times h + \frac{2 \times 0.06}{0.001} $$

First, calculate the surface tension term:

$$ \frac{2 \times 0.06}{0.001} = \frac{0.12}{0.001} = 120 \, \text{N/m}^2 $$

So the equation becomes:

$$ 1100 = 9800h + 120 $$

Now, solve for $$ h $$. Subtract 120 from both sides:

$$ 1100 - 120 = 9800h $$

$$ 980 = 9800h $$

Divide both sides by 9800:

$$ h = \frac{980}{9800} $$

Simplify the fraction:

$$ h = \frac{98}{980} = \frac{98 \div 98}{980 \div 98} = \frac{1}{10} = 0.1 \, \text{m} $$

Therefore, the depth of the bubble is 0.1 m.

Comparing with the options:

• A. 0.1 m
• B. 0.15 m
• C. 0.20 m
• D. 0.25 m

Hence, the correct answer is Option A.