Question

1 cc of water taken from the surface to the bottom of the lake having depth 100 metre. if Bulk modulus of water is 2.2×10^9 N/m^2 then decrease in the volume of the water will be​

Answer 1

If Bulk modulus of water is 2.2×10^9 N/m^2 , depth is 100 m and the volume of water is 1 cc then the decrease in the volume of the water will be 4.5 * 10⁻⁴ cc.

Explanation:

Given data:

The bulk modulus of water, K = 2.2 * 10⁹ N/m²

The initial volume of water, V = 1 cc

Depth of water, h = 100 m

Density of water, ρ = 1000 kg/m³

To find:

The decrease in the volume of the water

Solution:

Let the decrease in the volume of the water be denoted as “∆V”  cc.

We know,

The formula for Bulk Modulus is given as,

K = [Volumetric Stress] / [Volumetric Strain]

⇒ K = ∆P / [∆V/V]

Rearranging the formula

⇒ ∆V = [∆P * V] / K

⇒ ∆V = [(hρg)* V] / K

Substituting the given values, we get

⇒ ∆V = [(1000*100*10)* 1] / [2.2*10⁹]

⇒ ∆V = [10⁶ * 10⁻⁹ * 0.454]

⇒ ∆V = 0.454 * 10⁻³ cc

⇒ ∆V = 4.5 * 10⁻⁴ cc

Thus, the decrease in the volume of the water is 4.5 * 10⁻⁴ cc.

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Learn more:

Bulk modulus of water is 2 x 10^9 Nm^2-The change in pressure required to increase the density of water

by 0.1% is

(A) 2 x 10^9 Nm^2

(B) 2x10^8 Nm^2

(C) 2x10^6 Nm^2

(D) 2x10^4 Nm ^2

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Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

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