Question

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)`.

Answer 1

Correct Answer - `h=3.2,2.4,1.6,0.8,0;(dH)/(dt)=5 xx 10^(-3)sqrt(5H);Deltat=80(4-2sqrt(3))`
First resonance length
`v=n lambda`
`340=212.5 xx l_(1)`
`l_(1)=0.4m`
`x_(1)=3.6-0.4`
`x_(1)=3.2m`
Second resonance length
`V=nlambda_(2)implies 340=212.5xx(4l_(2))/(3)`
`l_(2)=1.2m`
`x_(2)=3.6-1.2`
`x_(2)=2.4m` Third resonance length
`v=n lambda implies 340=212.5xx(4)/(5)l_(3)`
`l_(3)=2mimpliesx_(3)=3.6-2impliesx_(3)=1.6m`
forth resonance length
`v=nimplies340=212.5xx(4l_(4))/(7)`
`l_(4)=2.8mimpliesx_(4)=3.6-2.8impliesx_(4)=0.8m`
Given that at any stage height of liquid above hole is H.
First we have to get time taken
velocity of efflux `v=sqrt(2gH)`
volume comingout per/sec. ` (dV)/ (dt )=A _(0 )sqrt (2gH)`
`(-AdH)/(dt)=A_(0)sqrt(2gH)`

`(dH)/(dt)=(A_(0))/(A)sqrt(2gH) " "implies(dH)/(dt)=(1xx10^(-3))/(2xx10^(-2))sqrt(2xx10xxH)`
`(dH)/(dt)=(pixx(1xx10^(-3))^(2))/(pixx(2xx10^(-2))^(2))sqrt(2xx10xxH)" "implies(dH)/(dt)=5xx10^(-3)sqrt(5H)`
For time interval
`(dV)/(dt)=A_(0)sqrt(2gH)`
`(-AdH)/(dt)=A_(0)sqrt(2gH)" "implies-AdH=A_(0)sqrt(2gH)dt`
`-(A)/(A_(0))underset(H)overset(x)(int)(dx)/(sqrt(x))=sqrt(2g)underset(0)overset(t)(int)dt" "implies-(A)/(A_(0))[sqrt(2x)]_(H)^(x)=sqrt(2g)t`
`(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(x)-sqrt(H)rfloor=t`
`t=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(H)-sqrt(x)rfloor`
for first resonance
`t_(1)=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(H)-sqrt(H_(1))rfloor`
for second resonance
`t_(2)=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(H)-sqrt(H_(2))rfloor" "so, "Deltat=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(H_(1))-sqrt(H_(2))rfloor`
`Deltat =(pi(2xx10^(-2))^(2))/(pi(1xx10^(-3))^(2))sqrt((2)/(10))(sqrt(3.2)-sqrt(2.4))`
`Deltat=400sqrt((1)/(5))(sqrt((16)/(5))-sqrt((12)/(5)))`
`Deltat=80(4-2sqrt(3))`sec.