Question

The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. The atmospheric pressure is 75 cm of mercury and the ratio of the density of mercury to that of lake water is 40/3. The depth of the lake is:

• A. 15 m
• B. 10 m
• C. 20 m
• D. 25 m

Hint:

Boyle’s law helps relate the pressure and volume of a gas at constant temperature. The pressure and volume are inversely proportional.

Answer 1

The Correct Option is C

The problem involves the relationship between pressure and volume for an air bubble as it rises to the surface of a lake. We can use Boyle's Law, which states that for a constant temperature, the product of pressure and volume remains constant, i.e., \( P_1V_1 = P_2V_2 \). Given:

\( V_2 = 2V_1 \) 

Let \( P_1 \) be the pressure at the bottom of the lake and \( P_2 \) be the pressure at the surface.

Atmospheric pressure \( P_2 = 75 \) cm Hg.

Density of mercury \( \rho_{Hg} = 40 \times \) density of water \( \rho_w \).

The pressure at the bottom is the sum of atmospheric pressure and the pressure due to water column:

\( P_1 = P_2 + h \rho_w g \)

Using Boyle’s Law:

\( P_1V_1 = P_2 \cdot 2V_1 \)

Therefore:

\( P_1 = 2P_2 \)

Substitute in the pressure equation:

\( 2P_2 = P_2 + h \rho_w g \)

\( P_2 = h \rho_w g \)

Convert \( P_2 \) from cm of Hg to pascals:

\( P_2 = 75 \) cm Hg \( = 75 \) cm \( \times 1333 \) Pascal/cm (since 1 cm Hg = 1333 Pa)

\( P_2 = 99975 \) Pascal

Using the conversion of mercury to water density:

\( \rho_w = \rho_{Hg} / (40/3) \)

\( \rho_w = 13600 \) kg/m\(^3\) / (40/3)

Calculate pressure in water as \( h \cdot \rho_w \cdot g = P_2 \):

\( \rho_w g = 99975 \) Pascal

Finally, solve for \( h \):

\( h = 99975 \) Pascal \(/ (1000 \times 9.81)\)

\( h \approx 20 \) meters

Thus, the depth of the lake is 20 meters.