The volume of an air bubble becomes 8 times the original volume in rising from the bottom of a lake to it's surface. If the barometric height is 0.76m of Hg, what is the depth of the lake?
(Density of Hg=13.6g/cm^3)
Answers
The height of the lake will be 72.35 m
Explanation :
Let P₁ P₂ and V₁ V₂ are pressure and volume at the bottom and top respectively,
Since temp remaining constant
P₁V₁ = P₂V₂
given, V₂ = 8V₁
=> P₁ = 8P₂
Hence difference in pressure at top and bottom:
P₁ - P₂
= 7P₂
Given density of Hg = 13.6 g/cm³ = 13600 kg/m³
Height of mercury column = 0.76 m
acceleration due to gravity = 9.8 m/s²
Hence P₂ = ρgh = 13600 x 9.8 x 0.76 = 101293
=> pressure difference = 7P₂ = 7 x 101293
= 709051
This pressure is equivalent to water pressure of height H, Hence
ρgH = 709051
=> 1000 x 9.8 x H = 709051
=> H = 72.35 m
Hence the height of the lake will be 72.35 m
Answer:
The height of the lake will be 72.35 m
Explanation :
Let P₁ P₂ and V₁ V₂ are pressure and volume at the bottom and top respectively,
Since temp remaining constant
P₁V₁ = P₂V₂
given, V₂ = 8V₁
=> P₁ = 8P₂
Hence difference in pressure at top and bottom:
P₁ - P₂
= 7P₂
Given density of Hg = 13.6 g/cm³ = 13600 kg/m³
Height of mercury column = 0.76 m
acceleration due to gravity = 9.8 m/s²
Hence P₂ = ρgh = 13600 x 9.8 x 0.76 = 101293
=> pressure difference = 7P₂ = 7 x 101293
= 709051
This pressure is equivalent to water pressure of height H, Hence
ρgH = 709051
=> 1000 x 9.8 x H = 709051
=> H = 72.35 m
Hence the height of the lake will be 72.35 m
Please mark me as the Brainliest