At what temperature will the speed of sound in air become double of its value of `0^@C`?

Explanation

We know that `v prop sqrt(T)`
So, `v_(2)/v_(1) = sqrt(T_(2)/T_(1))`
Let `T_(1) = T, T_(2) = T_(1) + t = T + t`
`v_(2)/v_(1) = sqrt((T+t)/T)`…….(i)
Putting `v_(2) = 2v_(1)`, in equation (i),
We have, `(2v_(1))/v_(1) = sqrt((T+1)/T)`
`2 = sqrt((T+t)/T) rArr 4 = (T +t)/T`
`4T = T + t`
t = 3T `rArr` As T = 273 K
So, `t = 3 xx 273 = 819 k = (819 - 273)^(@)C rArr t = 546^(@)C`
Therefore, at 546°C, speed of sound becomes double that at 0°C.