The density of water at `29^@C` is 996 kgm and an increase of pressure of `10^6" N/m"^2` diminishes the volume of 10^-3 mº3of water by `0.5 xx 10^(-6)m^3.` What is the velocity of sound in water at `29^@C.`

Answer

To find the velocity of sound in water at 29°C, we will use the relationship between bulk modulus, density, and the velocity of sound. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Density of water (ρ) = 996 kg/m³
- Increase in pressure (ΔP) = 10^6 N/m²
- Initial volume (V) = 10^-3 m³
- Change in volume (ΔV) = 0.5 × 10^(-6) m³

### Step 2: Calculate the bulk modulus (K)
The bulk modulus (K) is defined as:
\[ K = -\frac{ΔP}{\frac{ΔV}{V}} \]

Where:
- ΔP = change in pressure
- ΔV = change in volume
- V = initial volume

Substituting the values:
\[ K = -\frac{10^6}{\frac{0.5 \times 10^{-6}}{10^{-3}}} \]

Calculating the fraction:
\[ \frac{ΔV}{V} = \frac{0.5 \times 10^{-6}}{10^{-3}} = 0.5 \times 10^{-3} \]

Now substituting back into the equation for K:
\[ K = -\frac{10^6}{0.5 \times 10^{-3}} \]
\[ K = -\frac{10^6}{0.5} \times 10^{3} \]
\[ K = -2 \times 10^9 \text{ N/m²} \]

### Step 3: Calculate the velocity of sound (v)
The velocity of sound in a medium is given by the formula:
\[ v = \sqrt{\frac{K}{ρ}} \]

Substituting the values of K and ρ:
\[ v = \sqrt{\frac{2 \times 10^9}{996}} \]

Calculating the value:
\[ v = \sqrt{2.008016 \times 10^6} \]
\[ v \approx 1414.21 \text{ m/s} \]

### Final Answer
The velocity of sound in water at 29°C is approximately **1414.21 m/s**.

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