1q me itna hi diya hai
a sufficiently long organ pipe has a small hole at the bottom.initially the pipe is empty .water is poured into the pipe at the constant rate .the fundamental frequency of air column in d pipe
(a)cont increase
(b)first increase then constant
(c)cont decrease
(d)first decrease then constant
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13 Answers
13is the rate of the water coming out through the bottom is greater/lesser than the the rate of water being poured??
13as the length is decreasing...when water is being poured then...the freq will increase....
when the water going out through the hole then it will decrease again.....
13it will not become constant..
..if we continue pouring water after sometime the pipe will be filled with water[3]
if we stop pouring water then the length will increase again....and freq will decrease....
it will become constant only whan all the water will come out...[3]
13yes...thats the automatic choice as others are impossible....
bt i think the Q has some flaw...
1let is be filling at a rate= K (rate of volume increment)..
and the area of the hole be 'a'...the rate of outflow of water is √2gH as a function of height H filled.. so net rate of voulme increment = AdH/dt = (K-a*√2gH) where A is the area of the cylinder... now if H is filled by water, the free height is (L-H) where L is the total height of the cylinder.. now in that case the fundamental freq. f= v/ {2(L-H)} where v=velocity of sound in air..
now df/dt = [2v/{2(L-H)}2]*(dH/dt) {differentiating both sides}
now we know that dH/dt=(K-a*√2gH)/A...so we have
dF/dt = [2v/{2(L-H)}2]*(K-a*√2gH)/A...
looking at this expression we observe that dF/dt is positive only untill
K>a*√2gH or H<K2/(a*√2g)2...
so we observe that F first increases upto a certain limit of H and after that it remains constant as at this instant the rate of increment of volume becomes =0...
so B)
(here it becomes constant after wards becoz after a certain H the rate of volume increase=0 or no height increase or constant height, hence constant frequency)