Speed of sound at \(T_1 = 0^\circ C\) is \(V_0\) and at \(T_2 = \alpha^\circ C\) speed becomes \(2V_0\). Find \(\alpha\):
Show Hint
All formulas involving temperature (\(T\)) in gas laws (like ideal gas law, speed of sound) require the absolute temperature in Kelvin.
- \(819^\circ C\)
- \(918^\circ C\)
- \(546^\circ C\)
- \(1092^\circ C\)
The Correct Option is A
Solution and Explanation
To solve the problem of finding the temperature \(\alpha\) at which the speed of sound becomes \(2V_0\), we must first understand how the speed of sound in air changes with temperature.
The speed of sound in air is given by the formula:
\(V = V_0 \sqrt{\frac{T}{T_0}}\)
Where,
- \(V\) is the speed of sound at temperature \(T\).
- \(V_0\) is the speed of sound at 0°C (273 K).
- \(T\) is the temperature in Kelvin at which we want the speed of sound.
- \(T_0 = 273 \text{ K}\) is the reference temperature corresponding to 0°C.\)
Initially, at \(T_1 = 0^\circ C\), the speed of sound is \(V_0\).
At \(T_2 = \alpha^\circ C\), the speed of sound becomes \(2V_0\).
Given: \(V = 2V_0\),
Substitute this into the speed formula:
\(2V_0 = V_0 \sqrt{\frac{T_2}{273}}\)
Cancel \(V_0\) from both sides:
\(2 = \sqrt{\frac{T_2}{273}}\)
Square both sides to eliminate the square root:
\(4 = \frac{T_2}{273}\)
Rearrange to find \(T_2\):
\(T_2 = 4 \times 273\)
Calculating:
\(T_2 = 1092 \text{ K}\)
Convert Kelvin back to Celsius:
\(\alpha = T_2 - 273 = 1092 - 273 = 819^\circ C\)
Therefore, the correct answer is \(819^\circ C\).
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