Class 11 Physics Mechanical Properties of Solids Bulk Modulus

Bulk Modulus

• Bulk modulus is the ratio of hydraulic stress to the corresponding hydraulic strain.
• Denoted by ‘B’
• B = -p/(ΔV/V)
• Where p =hydraulic stress, ΔV/V = hydraulic strain

(-) ive signs show that the increase in pressure results in decrease in volume.

• S.I. Unit :- N/m2  or Pascal(Pa)
• B(solids) > B(liquids) >B(gases)

Compressibility

• Compressibility is the measure of compression of a substance.
• Reciprocal of bulk modulus is termed as ‘Compressibility’.
• Mathematically:
• k=1/B = - (1/p) (ΔV/V)
• It is denoted by ‘k’.
• k(solids)<k(liquids)<k(gases)

Problem:-The average depth of Indian Ocean is about 3000 m.

Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m–2. (Take g = 10 m s–2)

Answer: - The pressure exerted by a 3000 m column of water on the bottom layer

p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2

= 3 × 107 kg m–1 s-2

= 3 × 107 N m–2

Fractional compression ΔV/V, is

ΔV/V = stress/B = (3 × 107 N m-2)/ (2.2 × 109 N m–2)

= 1.36 × 10-2 or 1.36 %

Problem: The bulk modulus for water is 2.1GPa.Calculate the contraction in volume of 200ml of water is subjected to a pressure of 2MPa.

B=2.1GPa = 2.1 x109Pa.

V=200ml = 2x10-6ml.

P=2MPa = 200x106 Pa.

B=- (1/p) (ΔV/V) = ΔV = pV/B

= (2x106x200x10-6)/ = 2.1 x109

B=0.19ml

Problem: - 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.

Initial volume, V1 = 100.0l = 100.0 × 10–3 m3

Final volume, V2 = 100.5 l = 100.5 ×10–3 m3

Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3 m3

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

Bulk Modulus = Δp / ΔV/ V1 = Δp x V1/ ΔV

= (100x1.013 × 105x100x10-3)/0.5x10-3

=2.026x106 Pa

Bulk modulus of air= 1.0x105Pa

Therefore,

Bulk modulus of water/ Bulk modulus of air

=2.026x106/1.0x105 =2.026x104

This ratio is very high because air is more compressible than water.

Problem:-

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa

Density of water at the surface, ρ1 = 1.03 × 103 kg m–3

Let ρ2 be the density of water at the depth h.

Let V1 be the volume of water of mass m at the surface.

Let V2 be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = V1 - V2

=m (1/ ρ1 – 1/ ρ2)

Therefore, Volumetric strain = ΔV/ V1

=m (1/ ρ1 – 1/ ρ2) x ρ1/m

Therefore, ΔV/ V1 = (1- ρ1/ ρ2)     … (i)

Bulk modulus, B = p V1/ ΔV

ΔV/ V1 = p/B

Compressibility of water =1/B =45.8x10-11 Pa-1

Therefore, ΔV/ V1= 80x1.013x105x45.8x10-11 = 3.71x10-3    … (ii)

For equations (i) and (ii), we get:

1- ρ1 / ρ2 = 3.71 x10-3

ρ2 = 1.03x103/(1-(3.71x10-3)

=1.034x103kgm-3

Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.

Problem: - Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 105 Pa

Bulk modulus of glass, B = 37 × 109 Nm–2

Bulk modulus, B= p/ (ΔV/V)

Where,

ΔV/V = Fractional change in volume

Therefore,

ΔV/ V = p/B

=10x1.013x105

=2.73x10-5

Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.

Problem: - How much should the pressure on a litre of water is changed to compress it by 0.10%?

Volume of water, V = 1 L

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

Fractional change = ΔV/V =0.1/100x1 = 10-3

Bulk modulus, B= ρ/ ΔV/V

p=B x ΔV/V

Bulk Modulus of water, B = 2.2x109Nm-2

p=2.2x109x10-3

=2.2x106Nm-2

Therefore, the pressure on water should be 2.2 ×106 Nm–2.

Problem: - The average depth of Indian Ocean is about 3000 m. Calculate the

fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m–2. (Take g = 10 m s–2)

Answer:- The pressure exerted by a 3000 m column of water on the bottom layer

p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2

= 3 × 107 kg m–1 s-2

= 3 × 107 N m–2

Fractional compression ΔV/V, is

ΔV/V = stress/B = (3 × 107 N m-2)/ (2.2 × 109 N m–2)

= 1.36 × 10-2 or 1.36 %

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