Question

The increase in the pressure required to decrease the volume (\(\Delta V\)) of water is \(6.3 \times 10^7\) N/m². The percentage decrease in the volume is ______. (Bulk modulus of water = \(2.1 \times 10^9\) N/m².)

• A. 2 %
• B. 3 %
• C. 6 %
• D. 4 %

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Bulk modulus is a measure of a substance's resistance to uniform compression. It is defined as the ratio of volumetric stress (increase in pressure) to volumetric strain (fractional decrease in volume).
Step 2: Key Formula or Approach:
The formula for Bulk Modulus (\(B\)) is given by:
\(B = \frac{\Delta P}{\frac{\Delta V}{V}}\)
where \(\Delta P\) is the increase in pressure, \(\Delta V\) is the change in volume, and \(V\) is the original volume.
To find the percentage decrease in volume, we need to calculate \(\frac{\Delta V}{V} \times 100\).
Step 3: Detailed Explanation:
Given parameters:
Increase in pressure, \(\Delta P = 6.3 \times 10^7 \text{ N/m}^2\).
Bulk Modulus, \(B = 2.1 \times 10^9 \text{ N/m}^2\).
Rearranging the formula to solve for the fractional change in volume (\(\frac{\Delta V}{V}\)):
\(\frac{\Delta V}{V} = \frac{\Delta P}{B}\)
Substitute the known values into the equation:
\(\frac{\Delta V}{V} = \frac{6.3 \times 10^7}{2.1 \times 10^9}\)
\(\frac{\Delta V}{V} = 3 \times 10^{-2} = 0.03\)
Step 4: Final Answer:
The percentage decrease in volume is:
\(\text{Percentage decrease} = 0.03 \times 100% = 3%\).