Question

The ratio of speed of sound wave in steel to that of water is approximately. Given $Y = 2.0 \times 10^{11}$ Pa(for steel) and compressibility of water = 45.8 x $10^{-11}$ $Pa^{-1}$ and density of steel 7800 kg. $m^{-3}$

• A. 3.426
• B. 8.264
• C. 12.567
• D. 18.129

Answer 1

Answer: (A)

3.426

Answer 2

Solution:

1. Speed in Steel:

   $v_{\text{steel}} = \sqrt{\dfrac{Y}{\rho_{\text{steel}}}} = \sqrt{\dfrac{2.0 \times 10^{11}}{7800}} \approx \sqrt{2.564 \times 10^7} \approx 5064\, \text{m/s}$

2. Speed in Water:

   The bulk modulus of water, $K = \dfrac{1}{\text{compressibility}} = \dfrac{1}{45.8 \times 10^{-11}} \approx 2.183 \times 10^9\, \text{Pa}$.
   Using density of water $\rho_{\text{water}} \approx 1000\, \text{kg/m}^3$,
   $v_{\text{water}} = \sqrt{\dfrac{K}{\rho_{\text{water}}}} = \sqrt{\dfrac{2.183 \times 10^9}{1000}} \approx \sqrt{2.183 \times 10^6} \approx 1477\, \text{m/s}$.

3. Ratio of Speeds:

   $\dfrac{v_{\text{steel}}}{v_{\text{water}}} \approx \dfrac{5064}{1477} \approx 3.426$.

Core Explanation:
Calculated speeds using $v = \sqrt{\frac{\text{modulus}}{\rho}}$ for steel and water, then their ratio approximates 3.426.