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(B) The fundamental frequency $f$ of a closed organ pipe of length $L$ is given by $f = \frac{v}{4L}$,where $v$ is the speed of sound in air.As water

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first increases and then becomes constant

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(B) The fundamental frequency $f$ of a closed organ pipe of length $L$ is given by $f = \frac{v}{4L}$,where $v$ is the speed of sound in air.As water is poured into the pipe,the effective length $L$ of the air column decreases.Since $f \propto \frac{1}{L}$,as $L$ decreases,the fundamental frequency $f$ increases.However,the pipe has a small hole at the bottom. As the water level rises,the rate of water leaking out through the hole increases due to increased pressure.Eventually,the rate of water being poured in becomes equal to the rate of water leaking out,causing the water level to become constant.Once the water level becomes constant,the length $L$ of the air column remains constant,and therefore the fundamental frequency $f$ becomes constant.Thus,the frequency first increases and then becomes constant. The correct option is $B$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$K$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ sufficiently long closed organ pipe has a small hole at its bottom. Initially,the pipe is empty. Water is poured into the pipe at a constant rate. The fundamental frequency of the air column in the pipe:

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