A `3.6m` long vertical pipe resonates with a source of frequency `212.5 Hz` when water level is at certain height in the pipe. Find the height of water level (from the bottom of the pipe) at which resonance occurs. Neglect end correction. Now , the pipe is filled to a height `H (~~ 3.6m)`. A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of `H`. If the radii of the pipe and the hole are `2 xx 10^(-2)m` and `1 xx 10^(-3)m` respectively, Calculate the time interval between the occurance of first two resonances. Speed of sound in air `340m//s` and `g = 10m//s^(2)`.

Explanation

(i) 3.2m, 2.4m, 1.6m and 0.8m (ii)`(-dh)/(dt)=(1.11xx10^(-2))sqrt(H)` (b) 43 sec
Let `1_0` be the length of air column corresponding to thr fundamental frequency
`1_0=(lambda)/4` or `lambda=41_0`
Also `v=flambda`, Hence `f=(v)/(lambda)`
Then `(v)/(41_0)=212.5 or `1_0=(v)/(4(212.5))=(340)/(4(212.5))=0.4m
In closed pipe only odd harmonic are Now let `1_1,1_2,1_3,1_4,`etc. be the length corresponding to the `3^(rd)` harmonic
`3^(rd) harmonic, 3((v)/(41_1))=212.5 implies 1_1=1.2m , 5^(th) harmonic, 5((v)/(41_2))=212.5 implies1_2=2.0m`
`7^(th),harmonic 7((v)/(41_3))=212.5 implies 1_3=2.8m , 9^(th)harmonic,9((v)/(41_4))=212.5 implies 1_4=3.6m`
or heights of water level are (3.6-0.4)m,(3.6-1.2)m, (3.6-2.0)m, and (3.6-2.6)m
Therefore,heights of water level are 3.2m,2.4m,1.6m and 0.8m from bottom. Let A and a be the area of cross-sections of the pipe and hole respectively.
Then `A=pi(2xx10^(-2))^(2)=1.26xx10^(-3)m^(2) and a=pi(10^(-3))^(2)=3.14xx10^(-6)m^(2)`
Velocity of efflux `asqrt(2gH)`
Continuity equation at top of the tube and onifice `asqrt(2gH)=A((-dH)/(dt))`
Therefore rate of fall of water level in the pipe . `((-dH)/(dt))=a/Asqrt(2gH)`
Subsituting the values,we get `(-dH)/(dt)=(314xx10^(-6))/(1.26xx10^(-3))sqrt(2xx10xxH) implies (-dH)/(dt)=(1.11xx10^(-2))sqrtH`
Between first two resonance,the water level falls from 3.2m to 2.4m
`therefore (dH)/sqrtH=-11xx10^(-2)dt impliesint_(3.2)^(2.4)(dH)/(sqrtH)=-(1.11xx10^(-2))int_(0)^(t)dt`
`implies [sqrt(2.4)-sqrt(3.2)]=-(1.11xx10^(-2))t impliest=43sec`