A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water , figure. The water level in the tube is gradually lowerd. When the water level is 17cm below the open end, maximum intensity of sound in heard. If the room temperature is `20^(@)C`, calculate
(a) speed of sound in air at room temperature.
(b) speed of sound in air at `0^(@)C`
(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Explanation

Consider the diagram frequency of tuning v=512 Hz.

For observation of first maxima of intensity
(a) `L=(lambda)/(4)rArr=4L " "`
`" " v=vlambda=512xx4xx17xx10^(-2)`
`" " =384.16m//s`
(b) We know that `vpropsqrt(T)`
where temperature (T) is in kelvin.
`" " (v_(20))/(v_(0))=sqrt((273+20)/(273+0))=sqrt((293)/(273))`
`" " (v_(20))/(v_(0))sqrt(1.073)=1.03`
`" " v_(0)=(v_(20))/(1.03)=(384.16)/(1.03)=338 m//s`
(c) Resonance will be observed at 17 cm length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more danser than water.