Question

The volume of an air bubble becomes 8 times the original
volume in rising from the bottom of a lake to its surface. If the barometric height is 0.76 m of Hg (density of Hg is 13.6
g/cm3 and g=9.8 m/s2), what is the depth of the lake?
Aylinder hand​

Answer 1

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

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Explanation:

Answer 2

Answer:

72.35 m

Explanation:

Since volume of bubble become 8 times on reaching the surface.

Therefore pressure on surface of lake 1 / 8 times pf pressure at the bottom.

By Boyle's law pressure at bottom of the lake = 8 P [ P = pressure at surface. ]

= > Pressure due to column of liquid = 8 P - P = 7 P

P =  7 × 0.76 × 13.6 × 10³ × 9.8 N / m²

Now we know :

P = ρ h g

Density of mercury ρ = 13.6 × 10³ kg / m³

= > h = ( 7 × 0.76 × 13.6 × 10³ × 9.8 ) / ρ g

= > h =  ( 7 × 0.76 × 13.6 × 10³ × 9.8 ) / 10³ × 9.8

= > h = 72.35 m

Therefore , depth of the lake is 72.35 m.