Question:

If the pressure on a body is increased from \( 200 \) kPa to \( 250 \) kPa, the volume of the body decreases by \( 0.25% \). The compressibility of the material of the body (in \( m^2 N^{-1} \)) is:

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For compressibility calculations, use: \[ \beta = -\frac{\Delta V / V}{\Delta P} \] where \( \Delta V / V \) is the fractional volume change and \( \Delta P \) is the pressure difference.
Updated On: Jun 5, 2025
  • \( 2 \times 10^7 \)
  • \( 2 \times 10^{-7} \)
  • \( 5 \times 10^3 \)
  • \( 5 \times 10^{-8} \)
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The Correct Option is B

Solution and Explanation

Step 1: Compressibility Definition Compressibility \( \beta \) is defined as: \[ \beta = -\frac{\Delta V / V}{\Delta P} \] where: - \( \Delta V / V \) = Percentage volume change = \( 0.25% = 0.0025 \), - \( \Delta P = 250 - 200 = 50 \) kPa = \( 50 \times 10^3 \) Pa. Step 2: Calculating \( \beta \) \[ \beta = \frac{0.0025}{50 \times 10^3} \] \[ = \frac{2.5 \times 10^{-3}}{50 \times 10^3} \] \[ = 2 \times 10^{-7} \text{ m}^2 \text{N}^{-1} \] Conclusion Thus, the correct answer is: \[ 2 \times 10^{-7} \text{ m}^2 \text{N}^{-1} \]
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