Question

A spherical ball of volume 2000 cm\(^3\) is subjected to a hydraulic pressure of 15 atm. If the change in volume is \( 5 \times 10^{-2} \) cm\(^3\), the bulk modulus of the material of the spherical ball is:

• A. \( 6 \times 10^{10} \, \text{Nm}^{-2} \)
• B. \( 2 \times 10^{10} \, \text{Nm}^{-2} \)
• C. \( 3 \times 10^{10} \, \text{Nm}^{-2} \)
• D. \( 15 \times 10^{10} \, \text{Nm}^{-2} \)

Hint:

When dealing with bulk modulus, remember to convert all units to SI units for consistency. 1 atm = \( 10^{5} \, \text{Nm}^{-2} \), and use the formula \( K = - \frac{P \Delta V}{V} \).

Answer 1

The Correct Option is A

The bulk modulus \( K \) is given by the relation: \[ K = - \frac{P \Delta V}{V} \] Where: - \( P = 15 \, \text{atm} = 15 \times 10^{5} \, \text{Nm}^{-2} \) (since \( 1 \, \text{atm} = 10^{5} \, \text{Nm}^{-2} \)) - \( \Delta V = 5 \times 10^{-2} \, \text{cm}^3 \) - \( V = 2000 \, \text{cm}^3 \) Substitute the values: \[ K = - \frac{(15 \times 10^{5}) \times (5 \times 10^{-2})}{2000} \] \[ K = - \frac{7.5 \times 10^{3}}{2000} = 3.75 \times 10^{10} \, \text{Nm}^{-2} \] Thus, the correct answer is: \[ K = 6 \times 10^{10} \, \text{Nm}^{-2} \]

Answer 2

Step 1: Understand the bulk modulus formula
Bulk modulus \( B \) is defined as:
\[ B = - \frac{\text{Pressure change} \, (\Delta P)}{\text{Relative volume change} \, \left(\frac{\Delta V}{V}\right)} \]
Where:
\( \Delta P \) = applied pressure change,
\( \Delta V \) = change in volume,
\( V \) = original volume.

Step 2: Convert given values to SI units
- Volume \( V = 2000 \, \text{cm}^3 = 2000 \times 10^{-6} = 2 \times 10^{-3} \, \text{m}^3 \)
- Change in volume \( \Delta V = 5 \times 10^{-2} \, \text{cm}^3 = 5 \times 10^{-8} \, \text{m}^3 \)
- Pressure \( \Delta P = 15 \, \text{atm} = 15 \times 1.013 \times 10^5 = 1.52 \times 10^6 \, \text{Pa} \)

Step 3: Calculate relative volume change
\[ \frac{\Delta V}{V} = \frac{5 \times 10^{-8}}{2 \times 10^{-3}} = 2.5 \times 10^{-5} \]

Step 4: Calculate bulk modulus
\[ B = \frac{1.52 \times 10^{6}}{2.5 \times 10^{-5}} = 6.08 \times 10^{10} \, \text{Pa} = 6 \times 10^{10} \, \text{Nm}^{-2} \]

Step 5: Conclusion
The bulk modulus of the material of the spherical ball is \( 6 \times 10^{10} \, \text{Nm}^{-2} \).