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/m
^{2}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 × 10^{9} 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 × 10^{9} 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.

__Answer:-__

B=2.1GPa = 2.1 x10^{9}Pa.

V=200ml = 2x10^{-6}ml.

P=2MPa = 200x10^{6} Pa.

B=- (1/p) (ΔV/V) = ΔV = pV/B

= (2x10^{6}x200x10^{-6})/ = 2.1 x10^{9}

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 × 10

** Answer**:

Initial volume, V_{1} = 100.0l = 100.0 × 10^{–3} m^{3}

Final volume, V_{2} = 100.5 l = 100.5 ×10^{–3} m^{3}

Increase in volume, ΔV = V_{2 }– V_{1} = 0.5 × 10^{–3} m^{3}

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10^{5} Pa

Bulk Modulus = Δp / ΔV/ V_{1} = Δp x V_{1}/ ΔV

= (100x1.013 × 10^{5}x100x10^{-3})/0.5x10^{-3}

=2.026x10^{6} Pa

Bulk modulus of air= 1.0x10^{5}Pa

Therefore,

Bulk modulus of water/ Bulk modulus of air

=2.026x10^{6}/1.0x10^{5} =2.026x10^{4}

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 × 10^{3} kg m^{–3}?

__Answer:-__

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10^{5} Pa

Density of water at the surface, ρ_{1} = 1.03 × 10^{3} kg m^{–3}

Let ρ_{2} be the density of water at the depth h.

Let V_{1} be the volume of water of mass m at the surface.

Let V_{2} be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = V_{1} - V_{2}

=m (1/ ρ_{1} – 1/ ρ_{2})

Therefore, Volumetric strain = ΔV/ V_{1}

=m (1/ ρ_{1} – 1/ ρ_{2}) x ρ_{1}/m

Therefore, ΔV/ V_{1} = (1- ρ_{1}/ ρ_{2}) … (i)

Bulk modulus, B = p V_{1}/ ΔV

ΔV/ V_{1 }= p/B

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

Therefore, ΔV/ V_{1}= 80x1.013x10^{5}x45.8x10^{-11} = 3.71x10^{-3} … (ii)

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

1- ρ_{1 }/ ρ_{2} = 3.71 x10^{-3}

ρ_{2} = 1.03x10^{3}/(1-(3.71x10^{-3})

=1.034x10^{3}kgm^{-3}

Therefore, the density of water at the given depth (h) is 1.034 × 10^{3} kg m^{–3}.

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

__Answer:-__

Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 10^{5} Pa

Bulk modulus of glass, B = 37 × 10^{9} Nm^{–2}

Bulk modulus, B= p/ (ΔV/V)

Where,

ΔV/V = Fractional change in volume

Therefore,

ΔV/ V = p/B

=10x1.013x10^{5}

=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%?

__Answer:__

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 = 2.2x10^{9}Nm^{-2}

p=2.2x10^{9}x10^{-3}

=2.2x106Nm^{-2}

Therefore, the pressure on water should be 2.2 ×10^{6 }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 × 10^{9} 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 × 10^{7} kg m^{–1} s^{-2}

= 3 × 10^{7} N m^{–2}

Fractional compression ΔV/V, is

ΔV/V = stress/B = (3 × 10^{7} N m^{-2})/ (2.2 × 109 N m^{–2})

= 1.36 × 10^{-2} or 1.36 %

.