Sum

At what temperature will the speed of sound be double of its value at 0°C?

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#### Solution

Let the speed of sound *T*_{1} be *v*_{1},

where *T*_{1} = 0˚ C = 273 K.

Let *T*_{2}_{ }be the temperature at which the speed of sound (*v*_{2}) will be double its value at 0˚ C.

As per the question,*v*_{2} = 2*v*_{1}.

\[v \propto \sqrt{T}\]

∴

\[\frac{{v_2}^2}{{v_1}^2} = \frac{T_2}{T_1}\]

\[ \Rightarrow \frac{\left( 2 v_1 \right)^2}{{v_1}^2} = \frac{T_2}{273}\]

\[ \Rightarrow T_2 = 273 \times 4 = 1092 K\]

To convert Kelvin into degree celsius:

\[T_2 = \left( 273 \times 4 \right) - 273 = 819^\circ C\]

Hence, the temperature (*T*_{2}_{ }) will be 819˚ C

Concept: Wave Motion

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