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 ($B$) is defined as the ratio of the change in pressure (volumetric stress) to the volumetric strain. Percentage decrease in volume refers to $(\Delta V / V) \times 100$.

Step 2: Key Formula or Approach:
1. $B = \frac{\Delta P}{- \Delta V / V}$ (The negative sign indicates a decrease in volume with increasing pressure). 2. Volumetric strain $\frac{\Delta V}{V} = \frac{\Delta P}{B}$

Step 3: Detailed Explanation:
1. Calculate the volumetric strain: \[ \frac{\Delta V}{V} = \frac{6.3 \times 10^7}{2.1 \times 10^9} \] 2. Simplify the division: \[ \frac{\Delta V}{V} = \frac{6.3}{2.1} \times 10^{7-9} = 3 \times 10^{-2} \] 3. Convert to percentage: \[ \text{Percentage decrease} = \left( \frac{\Delta V}{V} \right) \times 100 = (3 \times 10^{-2}) \times 100 = 3 % \]

Step 4: Final Answer:
The percentage decrease in volume is 3 %.