Question

Pressure of \(2 \times 10^6\,\text{Pa}\) causes volume decrease of \(0.1\%\) in a material. Bulk modulus is:

• A. $2 \times 10^9\text{ Pa}$
• B. $2 \times 10^{10}\text{ Pa}$
• C. $1 \times 10^{10}\text{ Pa}$
• D. $5 \times 10^9\text{ Pa}$

Hint:

Always convert percentage changes into raw decimal fractions before substituting them into elastic moduli formulas ($0.1\% = 0.001 = 10^{-3}$). This avoids errors involving factors of $100$ in the final exponent.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The fundamental concept behind this problem is the elastic property of matter known as the Bulk Modulus of Elasticity (\(B\)).
Elasticity refers to the ability of a material to return to its original shape and size after the removal of deforming forces.
When a substance is subjected to a uniform external pressure (hydraulic stress) from all sides, its volume changes while its shape remains the same.
The Bulk Modulus specifically quantifies the resistance of a material to such uniform compression.
A higher value of Bulk Modulus indicates that the material is highly incompressible (like solids), whereas a lower value indicates high compressibility (like gases).
In physics, this is classified as a volumetric deformation, where the stress is the change in pressure and the strain is the fractional change in volume.
Step 2: Key Formula or Approach:
The mathematical definition of Bulk Modulus (\(B\)) is the ratio of volumetric stress to volumetric strain.
Volumetric Stress is equivalent to the applied excess pressure (\( \Delta P \)).
Volumetric Strain is the ratio of the change in volume (\( \Delta V \)) to the original volume (\( V \)).
The formula is expressed as:
\[ B = - \frac{\Delta P}{\frac{\Delta V}{V}} \]
The negative sign indicates that as the pressure increases (\( \Delta P>0 \)), the volume decreases (\( \Delta V<0 \)), making the ratio positive.
In most engineering and physics problems where we calculate the material constant, we deal with the magnitude:
\[ B = \frac{\Delta P}{\left| \frac{\Delta V}{V} \right|} \]
Step 3: Detailed Explanation:
Let us break down the numerical data provided in the question:
1. Applied Pressure (\( \Delta P \)): The question states that a pressure of \( 2 \times 10^{6} \) Pa is applied.
This represents the stress acting on the material. Units are in Pascals (Pa), which is \( \text{N/m}^{2} \).
2. Percentage Change in Volume: The material experiences a "volume decrease of 0.1%".
In terms of volumetric strain, a percentage must be converted into a raw fraction.
Mathematically, Percentage decrease = \( \left| \frac{\Delta V}{V} \right| \times 100 \).
So, \( 0.1 = \left| \frac{\Delta V}{V} \right| \times 100 \).
Dividing both sides by 100 gives the fractional change:
\[ \left| \frac{\Delta V}{V} \right| = \frac{0.1}{100} = 0.001 = 10^{-3} \]
3. Calculation of Bulk Modulus: Now, substitute the stress and the calculated strain into our formula.
\[ B = \frac{2 \times 10^{6}}{10^{-3}} \]
Applying the algebraic rule for exponents where \( \frac{x^{a}}{x^{b}} = x^{a-b} \), we have:
\[ B = 2 \times 10^{6 - (-3)} = 2 \times 10^{6 + 3} \]
\[ B = 2 \times 10^{9} \text{ Pa} \]
This result shows that the material is quite rigid, as it requires a significant amount of pressure to achieve even a small change in volume.
Step 4: Final Answer:
The Bulk Modulus of the material is calculated to be \( 2 \times 10^{9} \) Pa.