The temperature of an air bubble while rising from bottom to surface of a lake remains constant but diameter is doubled. If pressure on the surface is h meter of mercury column and relative density of mercury is ' ' then the depth of the lake is

Step-by-step solution

Since temperature remains constant, Boyle’s Law applies: P₁V₁ = P₂V₂ . The final diameter is double the initial diameter: D₂ = 2D₁ . For a sphere, V = 1 6 D^3 , so V₂ = 8V₁ . Surface pressure is P₂ = h _w g , and pressure at depth H is P₁ = P₂ + H _w g = (h + H) _w g . Substituting into Boyle’s Law: (h + H) _w g V₁ = h _w g 8V₁ . Canceling _w g V₁ gives h + H = 8h . Thus, H = 7h . Assuming h is 1 unit, option 7 is correct.