Q.If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1:1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?

[!TLDR] Since v is inversely proportional to sqrt(density) at fixed temperature, the day with the higher (1.1x) sound speed has less dense, i.e. moister, air; the density ratio works out to 1.21. [!ANSWER] Air is mo …

Given: on two days at the SAME temperature, the speeds of sound are in ratio $v_1:v_2=1:1.1$ (so $v_2=1.1v_1$, day 2 has the faster sound). By Laplace's formula at fixed temperature and pressure, $v=\sqrt{\gamma P/\rho}\propto\dfrac{1}{\sqrt{\rho}}$, so $\dfrac{v_2}{v_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}$. Squaring: $\dfrac{\rho_1}{\rho_2}=\left(\dfrac{v_2}{v_1}\right)^2=1.1^2=1.21$, i.e. $\rho_1=1.21\,\rho_2$ -- day 1's air is 21% denser than day 2's. Since moist air is LESS dense than dry air (section 8.5.5(c)), the DENSER air (day 1) must be the DRIER air, and the LESS dense air (day 2, with the higher sound speed) must be …

Apply $v\propto1/\sqrt{\rho}$ at fixed temperature/pressure to convert the given speed ratio into a density ratio, then use the fact that moist …

Concluding the FASTER-sound day is the drier one, by mistakenly assuming denser air conducts sound faster -- the opposite is true here, since it is elasticity-to-density ratio, and lower d …