Bulk Modulus of Elasticity Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Which one of the following is the correct relationship between Young’s modulus (Y), Bulk modulus (B), and Poisson’s ratio (σ)?

1. Y = 2B(1-2σ)
2. Y = 3B(1-2σ)
3. 2Y = 3B(1-2σ)
4. Y = 3σ(1-2B)

Ans. The correct answer is b. Y = 3B(1-2σ)

Explanation: The relationship between Young’s modulus (Y), Bulk modulus (B), and Poisson’s ratio (σ) is given by Y =3B(1-2σ)

Ques. The Bulk Moduli of 3 fluids, that is, A, B, and C are K_A, K_B, and K_C respectively. If K_A > K_B > K_C, then which liquid will have the maximum compressibility?

1. Liquid A
2. Liquid B
3. Liquid C
4. All will have the same compressibility

Ans. The correct answer is c. Liquid C

Explanation: The reciprocal of the bulk modulus is known as Compressibility. The more the bulk modulus less will the compressibility.

Therefore liquid C has less bulk modulus, therefore its compressibility is higher.

Ques. Which one of these won’t have the same unit as the others?

1. Force
2. Pressure
3. Bulk modulus
4. Stress

Ans. The correct answer is a. Force

Explanation: The unit of force is Newton, while the unit of pressure, bulk modulus, and stress are Pascal (Pa), or N/m2

Ques. Which among the following is the unit of compressibility?

1. m²/N
2. m/N²
3. m/N
4. m²/N²

Ans. The correct answer is a. m²/N

Explanation: Compressibility is the reciprocal of the bulk modulus. The unit of the bulk modulus is N/m²

Therefore the unit of compressibility is m²/N.

Ques. For an incompressible fluid, what will be the value of the Bulk Modulus of elasticity?

1. Unity
2. Infinity
3. Zero
4. Very low

Ans. The correct answer is b. Infinity

Explanation: The bulk modulus (B) is reciprocal of the compressibility (β) i.e.

B = 1/β

For an incompressible fluid, β = 0.

Therefore, B = 1/0 = infinity

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Short Answers Questions [2 Marks Questions]

Ques. Define bulk stress and bulk strain.

Ans. Bulk stress is defined as the force acting perpendicular to the surface of the object per unit area of the object. i.e.

Buk stress = Pressure = Force/Area

When a body is under bulk stress then bulk strain is defined as the ratio of the change in volume to the original volume. i.e. 

Bulk strain = Change in volume (ΔV) / Original volume (V)

Ques. Define the Bulk modulus of elasticity.

Ans. When Hooke’s law is obeyed, an increase in pressure i.e. bulk stress produces a proportional bulk strain i.e. fractional change in volume. The corresponding elastic modulus i.e. the ratio of bulk stress to bulk strain is called the bulk modulus of elasticity.

Ques. What is the formula of the Bulk modulus of elasticity?

Ans. The formula of bulk modulus of elasticity is given by

B = – Δp/(ΔV/V)

Where

• Δp is the change in pressure
• ΔV is the change in volume
• V is the original volume

Ques. Define the modulus of elasticity.

Ans. The ratio of the stress to strain is called the modulus of elasticity. There are three types of modulus of elasticity

• Young’s modulus of elasticity
• Shear modulus of elasticity or modulus of rigidity
• The bulk modulus of rigidity

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Long Answers Questions [3 Marks Questions]

Ques. A spherical ball contracts in volume by 0.05%, when subjected to a normal uniform pressure. Calculate the volume strain produced in the spherical ball.

Ans. Let V be the initial volume of the spherical ball

Given, the ball contracts in volume by 0.05%. Therefore

Change in volume, ΔV = 0.05% of V = (0.05/100)V

Volume strain or Bulk strain produced = ΔV/V = (0.05/100)V/V= 5 x 10^-4

Ques. When the temperature of the gas is 20 ^oC and pressure is changed from P₁ = 1.01 x 10⁵ Pa to P₂ = 1.165 x 10⁵ Pa, then the volume changes by 10%. Find the Bulk modulus of the gas.

Ans. Change in pressure, Δp = P₂ - P₁ = 1.165 x 10⁵ - 1.01 x 10⁵

⇒ Δp = 0.155 x 10⁵ Pa

Let V be the original volume of the gas, then fractional change in volume is given by

ΔV/V = 10% = 10/100 = 0.1

The bulk modulus of elasticity is given by

B = Δp/(ΔV/V)

⇒ B = (0.155 x 10⁵)/0.1 = 1.55 x 10⁵ Pa

Ques. If the bulk modulus of lead is 8.0 x 10⁹ N/m² and the initial density of the lead is 11.4 g/cc, then what is the density of the lead under the pressure of 2 x 10⁸ N/m²?

Ans. Given

• The bulk modulus of lead, B = 8.0 x 10⁹ N/m²
• The initial density of lead, ρi = 11.4 g/cc
• Change in pressure, Δp = 2 x 10⁸ N/m²

Let Vi be the initial volume, then the bulk modulus of elasticity is given by

B = - Δp/(ΔV/V_i)

⇒ ΔV/V_i = - Δp/B = - (2 x 10⁸)/(8.0 x 10⁹) = - 1/40

⇒ ΔV = - V_i/40

⇒ V_f - V_i = - V_i/40

Where V_f is the final volume

⇒ V_f = - 39V_i/40

⇒ |V_i/V_f| = 40/39

Let M be the mass of the lead, then

M = ρ_iV_i = ρ_fV_f

Where ρ_f is the final density of the lead

ρ_f = ρ_iV_i/V_f = 11.4 x (40/39) = 11.7 g/cc

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Very Long Answers Questions [5 Marks Questions]

Ques. A hydraulic press contains 0.25 m³ (250 L) of oil. Find the decrease in the volume of the oil when it is subjected to a pressure increase Δp = 1.6 x 10⁷ Pa. The bulk modulus of oil is B = 5.0 x 10⁹ Pa and its compressibility is K = 1/B = 20 x 10^-6 Pa^-1

Ans. Given

• The volume of the oil, V = 0.25 m³
• Change in pressure, Δp = 1.6 x 10⁷ Pa
• The bulk modulus of oil, B = 5.0 x 10⁹ Pa

We have

Bulk modulus, B = - Δp/(ΔV/V) = - (Δp x V)/ΔV

Where ΔV is the change in volume.

Therefore, ΔV = - (Δp x V)/B

⇒ ΔV = - (1.6 x 10⁷ x 0.25)/5.0 x 10⁹ = - 0.8 x 10^-3 m³

Since 1 m³ = 10³ L

⇒ ΔV = - 0.8 x 10^-3 x 10³ L = - 0.8 L

The negative sign shows that there is a decrease in the volume of the oil. Hence the decrease in the volume of oil is 0.8 L

Ques. The average depth of the Indian Ocean is 3000 m. Calculate the fraction compression, ΔV/V of water at the bottom of the ocean, given the bulk modulus of water is 2.2 x 10⁹ N/m².

Ans. The bulk modulus of elasticity is given by

B = - Δp/(ΔV/V)

Where

• Δp is change in pressure
• ΔV is chage in volume
• V is original volume

Therefore

|- ΔV/V| = Δp/B

⇒ΔV/V = Δp/B

Now change in pressure due to height h = 3000 m column of water at the bottom is given by

Δp = ρgh = 1000 x 10 x 3000 = 3 x 10⁷ Pa

Therefore, the fraction compression of water

ΔV/V = Δp/B = (3 x 10⁷)/(2.2 x 10⁹) = 1.36 x 10^-2 or 1.36%.

Ques. 1 cc of water is taken from the surface to the bottom of a lake having a depth of 100 m. If the bulk modulus of water is 2.2 x 10⁹ N/m² then find the decrease in volume of water.

Ans. The bulk modulus of elasticity is given by

B = - Δp/(ΔV/V)

Where

• Δp is change in pressure
• ΔV is chage in volume
• V is original volume

Therefore

|-ΔV| = (Δp x V)/B

⇒ ΔV = (Δp x V)/B

Now change in pressure due to height h = 100 m column of water at the bottom is given by

Δp = ρgh = 1000 x 10 x 100 = 10⁶ Pa

Therefore, the decrease in the volume of water is given by

ΔV/V = (Δp x V)/B = (10⁶ x 1 cc)/(2.2 x 10⁹) = 4.4 x 10^-4 cc

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