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JAMB Physics 2023 Paper

Question : 19

Total: 65

Solution:

Concept:

As the bubble rises, the external pressure decreases, so its volume increases according to Boyle’s law at constant temperature.

Explanation:

Let

r_1 = 4.5\ \mathrm{cm}

be the initial radius at a depth of

12\ \mathrm{m}

The total pressure at that depth is

P_1 = 12 + 10.34 = 22.34\ \mathrm{m}

of water.

At the surface, the pressure is

P_2 = 10.34\ \mathrm{m}

of water.

The volume of a spherical bubble is

V = \frac{4}{3}\pi r^3

Using Boyle’s law:

P_1 V_1 = P_2 V_2

Substitute the values:

22.34 \times \frac{4}{3}\pi (4.5)^3 = 10.34 \times \frac{4}{3}\pi r_2^3

Cancel

\frac{4}{3}\pi

from both sides:

22.34 \times (4.5)^3 = 10.34 \times r_2^3

r_2^3 = \frac{22.34 \times 91.125}{10.34} = 196.88

r_2 = \sqrt[3]{196.88} = 5.82\ \mathrm{cm}

Answer:

5.82\ \mathrm{cm}

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