Question

find the change in volume which 1m³ of water will undergo when taken from the surface to the bottom of a lake 100 m deep. Given volume elasticity of water is 22000 atmosphere​

Answer 1

Answer:

Explanation:

Volume   V=1 m  

3

 

Bulk modulus   B=20000 N/m  

2

 

Excess pressure at 1 km depth,     ΔP=ρgh=1000×10×1000=10  

7

 N/m  

2

 

Using    B=  

ΔV

−VΔP

​

 

Or     20000=  

ΔV

−1×10  

7

 

​

 

ΔV=−2×10  

−3

 m  

3

Answer 2

We have to find the change in volume which 1m³ of water will undergo when taken from the surface to the bottom of a lake 100 m deep.

Given volume elasticity of water is 22000 atm.

pressure at the bottom of a lake 100m deep, P = P₀ + ρgh

here, P₀ = 1.013 × 10⁵ N/m² [ atmospheric pressure ]

ρ = density of water = 1000 kg/m³

h = 100 m

now, P = 1.013 × 10⁵ + 1000 × 10 × 100

= 1.013 × 10⁵ + 10 × 10⁵

= 11.103 × 10⁵ N/m²

volume elasticity (bulk modulus), β = 22000 atm

= 22000 × 1.013 × 10⁵

= 0.22286 × 10¹⁰ N/m²

and initial volume of water, V = 1 m³

now using formula,

$\quad\beta=\frac{P}{\frac{\Delta V}{V}}$

⇒0.22286 × 10¹⁰ = 11.1013 × 10⁵/(∆V/1)

⇒∆V = 11.013/(0.22286) × 10¯⁵

= 49.4167 × 10¯⁵

= 4.94167 × 10¯⁴ m³ ≈ 5 × 10¯⁴ m³

Therefore the change in volume is 5 × 10¯⁴ m³ (approximately)