The speed of sound in hydrogen is `x xx 332` m/s at NTP if density of hydrogen is `(1)/(4)` the of that of air. Find the value of x if speed of sound in air is 332 m/s at NTP.
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Explanation

To solve the problem, we need to find the value of \( x \) in the speed of sound in hydrogen given that the speed of sound in air is 332 m/s and the density of hydrogen is \( \frac{1}{4} \) that of air.

### Step-by-Step Solution:

1. **Understand the relationship between speed of sound and density**:
The speed of sound \( v \) in a medium is given by the formula:
\[ v = \sqrt{\frac{\gamma}{\rho}} \]
where \( \gamma \) is the adiabatic index (ratio of specific heats) and \( \rho \) is the density of the medium.

2. **Set up the ratio of speeds**:
For hydrogen and air, we can write:
\[ \frac{v_{\text{air}}}{v_{\text{hydrogen}}} = \sqrt{\frac{\rho_{\text{hydrogen}}}{\rho_{\text{air}}}} \]

3. **Substitute the given values**:
We know that the density of hydrogen is \( \frac{1}{4} \) that of air:
\[ \rho_{\text{hydrogen}} = \frac{1}{4} \rho_{\text{air}} \]
Therefore, substituting this into the equation gives:
\[ \frac{v_{\text{air}}}{v_{\text{hydrogen}}} = \sqrt{\frac{\frac{1}{4} \rho_{\text{air}}}{\rho_{\text{air}}}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

4. **Relate the speeds**:
From the ratio, we can express the speed of sound in hydrogen:
\[ v_{\text{air}} = \frac{1}{2} v_{\text{hydrogen}} \]
Rearranging gives:
\[ v_{\text{hydrogen}} = 2 v_{\text{air}} \]

5. **Substitute the speed of sound in air**:
We know that \( v_{\text{air}} = 332 \, \text{m/s} \):
\[ v_{\text{hydrogen}} = 2 \times 332 = 664 \, \text{m/s} \]

6. **Express the speed of sound in hydrogen in terms of \( x \)**:
The problem states that the speed of sound in hydrogen is \( x \times 332 \, \text{m/s} \). Therefore, we have:
\[ 664 = x \times 332 \]

7. **Solve for \( x \)**:
Dividing both sides by 332 gives:
\[ x = \frac{664}{332} = 2 \]

### Final Answer:
The value of \( x \) is \( 2 \).

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