Question

The wave velocity of ultrasound in soft tissues is 1540 m/s and the impedance offered by it is \( 1.63 \times 10^6 \, \text{kg/m}^2\text{s} \). What is the density of the soft tissue?

• A. \( 0.1058441 \, \text{kg/m}^3 \)
• B. \( 10.58441 \, \text{kg/m}^3 \)
• C. \( 1058.441 \, \text{kg/m}^3 \)
• D. \( 105844.1 \, \text{kg/m}^3 \)

Hint:

Remember: Acoustic impedance \( Z = \rho \cdot c \). Ensure all units are consistent—velocity in m/s, impedance in \( \text{kg/m}^2\text{s} \), and density in \( \text{kg/m}^3 \).

Answer 1

The Correct Option is C

Understanding the Problem:

• Ultrasound wave velocity (\( v \)) in soft tissues is given as 1540 m/s.
• Acoustic impedance (\( Z \)) of the soft tissue is \( 1.63 \times 10^6 \, \text{kg/m}^2\text{s} \).
• We need to find the density (\( \rho \)) of the soft tissue.

Formulae:

• The acoustic impedance (\( Z \)) is related to the density (\( \rho \)) and wave velocity (\( v \)) by: \[ Z = \rho \times v \]

Rearranging the Formula:

To find density (\( \rho \)), we rearrange the formula: \[ \rho = \frac{Z}{v} \]

Substituting Given Values:

\[ \rho = \frac{1.63 \times 10^6 \, \text{kg/m}^2\text{s}}{1540 \, \text{m/s}} \]

Calculating the Density:

\[ \rho = \frac{1.63 \times 10^6}{1540} \approx 1058.441 \, \text{kg/m}^3 \]

Final Answer:

The density of the soft tissue is \( 1058.441 \, \text{kg/m}^3 \).