An air bubble of volume rises from the bottom of a lake 40 m deep to the surface at a temperature of . The atmospheric pressure is , the density of water is and . [2023]

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Solution

Q.
An air bubble of volume ${1 cm}^{3}$ rises from the bottom of a lake 40 m deep to the surface at a temperature of $12 ° C$ . The atmospheric pressure is $1 \times 10^{5} Pa$ , the density of water is $1000 {kg/m}^{3}$ and $g = 10 {m/s}^{2}$ . There is no difference of the temperature of water at the depth of 40 m and on the surface. The volume of the air bubble when it reaches the surface will be [2023]

1 2 ${cm}^{3}$

2 4 ${cm}^{3}$

3 5 ${cm}^{3}$

4 3 ${cm}^{3}$

Ans.
(3)

$P = P_{0} + ρ g h = 10^{5} Pa + 10^{3} \times 10 \times 40 = 5 \times 10^{5} Pa$

At T is constant, $P V = P_{0} V_{0}$

$\Rightarrow 5 \times 10^{5} Pa \times 1 {cm}^{3} = 10^{5} Pa \times V_{0}$ $\Rightarrow V_{0} = 5 {cm}^{3}$

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