The speed of sound in air is `332 ms^(-1)` at 0°C. At what temperature will the speed become half of that at `0^(@)` C ?

Explanation

We are given, `v_(1) = 332 ms^(-1)` and `v_(2) = 166 ms^(-1)`
`T_(1) = 0 + 273 = 273 K, T_(2)` = ?
We are to calculate, `T_(2)`
We know that, `v prop sqrt(T)`
Using relation, `v_(1)/v_(2) = sqrt(T_(1)/T_(2))` we have `T_(2) = (v_(2)/v_(1))^(2) xx T_(1)`
`T_(2) =(166/332)^(2) xx 273` or `T_(2)=1/4 xx 273 = 68.25 K`