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

The acoustic impedance \( Z \) of a medium is given by: \[ Z = \rho \cdot c \] where: \( Z = 1.63 \times 10^6 \, \text{kg/m}^2\text{s} \), \( c = 1540 \, \text{m/s} \), \( \rho \) is the density in \( \text{kg/m}^3 \). Rearranging the formula to solve for density: \[ \rho = \frac{Z}{c} = \frac{1.63 \times 10^6}{1540} \] \[ \rho = 1058.441558 \, \text{kg/m}^3 \approx \boxed{1058.441 \, \text{kg/m}^3} \]