Standing waves are set up in some organ pipes (open or closed) with maximum pressure amplitude $P_0$  inside them. Pressure amplitude at a distance x  from open end of the pipes is  $P_x.$ Column (I) gives values of  $P_x$  for select values of x  Match them with possible pipe length and frequency of excitation given in Column (II). Speed of sound in air  $=320m/s.$

Column – I		Column – II
A)	$x=8cm,P_x=P_0$	P)	Length $32cm,f=500Hz$
B)	$x=8cm,P_x=\frac{P_0}{\sqrt{2}}$	Q)	Length $8cm,f=500Hz$
C)	$x=16cm,P_x=0$	R)	Length $48cm,f=500Hz$
D)	$x=48cm,P_x=P_0$	S)	Not possible

$\lambda =\frac{v}{f}=\frac{320}{500}=0.64cm=64cm.$

If the pipe is closed, it’s possible lengths are  $\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4}\mathrm{....}=16,48,80cm\mathrm{.....}$

With  $l=48cm,$ the pipe oscillates as shown. For  $x=8,$
$P_x=P_0\sin (\frac{2\pi }{64}\times 8)=\frac{P_0}{\sqrt{2}}.$ 
At  $x=48,P_x=P_0$
If the pipe is open, its possible lengths are   $\frac{\lambda }{2},\lambda ,\frac{3\lambda }{2}\mathrm{...}=32cm,64cm,96cm\mathrm{....}$
With  $l=32cm$  and  $\lambda =64cm,$ the pipe is in its fundamental mode.

For  $x=8cm,$
 $P_x=P_0\sin (\frac{2\pi }{64}\times 8)=\frac{P_0}{\sqrt{2}}.$
At   $\frac{\lambda }{4}=16cm$ from open end there will always be an antinode. Therefore, option (C) in column (I) will match with only (S) in column (II). Similar is the case for (A).