Question

how much pressure is needed to compress a sample of water by 0.4%

Answer 1

Answer:

original volume of water , V = 1 L

Let change in volume of water is ∆V

A/C to question,

% change in volume of water = ∆V/V × 100

0.1 = ∆V/V × 100

∆V = 0.001

from data, bulk modulus of water,\betaβ = 2.2 × 10^9 N/m²

use formula, \beta=\frac{P}{\frac{\Delta{V}}{V}}β=

V

ΔV

P

so, P = \beta\left(\frac{\Delta{V}}{V}\right)β(

V

ΔV

)

P = 2.2 × 10^9 × 0.001

P = 2.2 × 10^6 N/m²

hence, the pressure on water should be 2.2 × 10^6 N/m²

Answer 2

Explanation:

Volume of water, V=1L

It is given that water is to be compressed by 0.10%.

∴ Fractional change, △V/V=(0.1/100)×1=10

−3

Bulk modulus, B=P/(△V/V)

P=B×(△V/V)

Bulk modulus of water, B=2.2×10

9

Nm

−2

P=2.2×10

9

×10

−3

=2.2×10

6

Nm

−2

Hence, the pressure on water should be 2.2×10

6

Nm

−2

Answer By....

hope you understand ...