The speed of sound as measured by a student in the laboratory on a winter day is `340 m/s` when the room temperature is `17^@C`. What speed will be measure another student repeating the experiment on a day when the room temperature is `32@C` ?

Explanation

To find the speed of sound at a different temperature, we can use the relationship that the speed of sound is proportional to the square root of the absolute temperature. Here’s a step-by-step solution:

### Step 1: Convert the temperatures to Kelvin
The temperatures given in Celsius need to be converted to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

- For \( T_1 = 17°C \):
\[ T_1 = 17 + 273.15 = 290.15 \, K \]

- For \( T_2 = 32°C \):
\[ T_2 = 32 + 273.15 = 305.15 \, K \]

### Step 2: Use the formula for the speed of sound
The speed of sound \( V \) is proportional to the square root of the absolute temperature:
\[ \frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} \]

### Step 3: Substitute the known values
We know:
- \( V_1 = 340 \, m/s \)
- \( T_1 = 290.15 \, K \)
- \( T_2 = 305.15 \, K \)

Now we can substitute these values into the equation:
\[ V_2 = V_1 \cdot \sqrt{\frac{T_2}{T_1}} \]
\[ V_2 = 340 \cdot \sqrt{\frac{305.15}{290.15}} \]

### Step 4: Calculate the ratio of temperatures
First, calculate the ratio:
\[ \frac{T_2}{T_1} = \frac{305.15}{290.15} \approx 1.050 \]

### Step 5: Calculate the square root of the ratio
Now, calculate the square root:
\[ \sqrt{1.050} \approx 1.0247 \]

### Step 6: Calculate \( V_2 \)
Now substitute back to find \( V_2 \):
\[ V_2 = 340 \cdot 1.0247 \approx 348.00 \, m/s \]

### Final Result
Thus, the speed of sound measured on the day when the room temperature is \( 32°C \) is approximately:
\[ V_2 \approx 348 \, m/s \]