Q69 / 90 · JEE Main 2023 (08 Apr Shift 1) · Physics

An air bubble of volume 1 cm3 rises from the bottom of a lake 40 m deep to the surface at a temperature of 12°C. The atmospheric pressure is  1×105 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 air bubble when it reaches the surface will be

  • A
    2 cm3
  • B
    3 cm3
  • C
    4 cm3
  • Correct answer
    D
    5 cm3

Solution

The pressure at the bottom of the lake can be calculated as follows:

Pbottom= P0 + ρgh=105+1000 × 10 × 40 Pa=5×105 Pa 

As the temperature of water remains constant, it can be written that

PbottomVbottom=PtopVtop   ...(1)

Substitute the values of the known parameters into equation (1) to calculate the required volume of the bubble when it reaches the surface.

5×105×1=1×105×Vtop⇒Vtop=5×105×11×105 cm3= 5 cm3

Hence, the correct answer is option: 'D'.

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