Calculate the temperature at which the speed of sound will be two times its value at $0\,^oC$. (in $,K$)

Use the formula $v=\sqrt{\frac{\gamma P}{\rho}}$ to explain why the speed of sound in air:
$(a)$ is independent of pressure,
$(b)$ increases with temperature,
$(c)$ increases with humidity.

$A$ tuning fork of frequency $220 \,Hz$ produces sound waves of wavelength $1.5 \,m$ in air at $N.T.P$. The increase in wavelength when the temperature of air is $27^{\circ} C$ is nearly $\left(\sqrt{\frac{300}{273}}=1.05\right)$ (in $\,m$)

$A$ sound wave is travelling with a frequency of $50 \,Hz$. The phase difference between two points in the path of the wave is $\frac{\pi}{3} \,rad$. The distance between those two points is (Velocity of sound in air $=330 \,m/s$) (in $m$)

The speed of sound in oxygen $(O_2)$ at a certain temperature is $460 \,ms^{-1}$. The speed of sound in helium $(He)$ at the same temperature will be (assume both gases to be ideal): (in $\,ms^{-1}$)

The speed of sound in air is independent of its: