Answer 1

Let the speed of sound T₁ be v₁,
where T₁ = 0˚ C = 273 K.
Let T₂ be the temperature at which the speed of sound (v₂) will be double its value at 0˚ C.
As per the question,
v₂ = 2v₁.

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

∴​

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

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

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

To convert Kelvin into degree celsius:

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

Hence, the temperature (T₂ ) will be 819˚ C