If an organ pipe is sounded with a tuning fork of frequency 256 Hz, the resonance occured at 35 cm and 105 cm, then the velocity of sound is?

It is an organ pipe with one closed end.

For a close ended organ pipe: The fundamental (first harmonic) needs to have an node at the close end since the air cannot move and an antinode at the open end.

#λ/4 = L_1 + e# …... (1)
where #e# is the end correction.

Another resonance occurs at #L_2=105cm#
For a closed end organ pipe this is the 3rd harmonic resonating as there is no 2nd harmonic for closed ended pipes.

Noi abbiamo

#(3λ)/4 = L_2 + e # …... (2)

Subtracting equations (1) from (2) we get

#λ/2 = (L_2 – L_1) # …... (3)

Using the expression

#v=flambda#
where #v# is velocity of sound, #f# is frequency and #lambda# is the wavelength.

otteniamo

#v = 2f (L_2 - L_1) #

Inserting given values we get

#v=2xx256(105-35)#
#=>v=2xx256(105-35)#
#=>v=35840cms^-1#