Q. The temperature at which the speed of sound in air becomes double of its value at $0℃$ is
A) $273\ K$
B) $546\ K$
C) $1092\ K$
D) $0\ K$
Explanation
(c) $v \propto \sqrt{T} \Rightarrow \sqrt{\frac{T_{2}}{T_{1}}} = \frac{v_{2}}{v_{1}} \Rightarrow T_{2} = T_{1}\left( \frac{v_{2}}{v_{1}} \right)^{2}$
$\Rightarrow T_{2} = 273 \times 4 = 1092K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.