The volume of an air bubble is doubled as it rises
from the bottom of lake to its surface. The atmospheric
pressure is 75 cm of mercury. The ratio of density of
mercury to that of lake water is40/3
. The depth of
the lake in metre is
Answers
Answer:
Given:
ratio of the volume of the air bubble at the top to the bottom, V'/V = 2,
=> V' = 2V
ratio of the density of the mercury to that of the lake water, ρ'/ρ = 40/3
=> ρ' = (40/3) ρ
Now, the pressure exerted at the surface is equal to 75 cm of mercury, so here depth = 0.75 m
the pressure exerted at the bottom will be P at depth = pressure due to water column h + atmospheric pressure, here, h is the depth of the lake
so, the equation would be P' V' = P V
where P = hρg + atmospheric pressure
h is equal to 10 m
Explanation:
follow me
like my answer
Answer:
Explanation:
Given:
ratio of the volume of the air bubble at the top to the bottom, V'/V = 2,
=> V' = 2V
ratio of the density of the mercury to that of the lake water, ρ'/ρ = 40/3
=> ρ' = (40/3) ρ
Now, the pressure exerted at the surface is equal to 75 cm of mercury, so here depth = 0.75 m
the pressure exerted at the bottom will be P at depth = pressure due to water column h + atmospheric pressure, here, h is the depth of the lake
so, the equation would be P' V' = P V
where P = hρg + atmospheric pressure
h is equal to 10 m