A certain organ pipe, open at both ends, produces a fundamental frequency of 288 Hz in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency tHe will it produce

773.25 Hz

Explanation:

Concept : In an open organ pipe in fundamental mode of vibration

wave length of wave λ = 2L

where L  is length of  the pipe

frequency   = velocity of sound / λ

Given values: fundamental frequency = 288 Hz

fluid is air. velocity of sound = 340 m/s

⇒ 288 = 340/2L

⇒L = 59.02 cm

The point to be noted is if the pipe is filled with helium initially at the same temperature, there would be change in the sound velocity .Then,  frequency of note produced will also be changed .

We know that velocity of sound is inversely proportional to  square root of molar mass of gas

velocity of sound in air / velocity of sound in helium = Square root of (Molar mass of Helium/ molar mass of air)

Now, frequency  = velocity of sound / λ

= 912.75 / (2 x 0.5902)

= 773.25 Hz

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.

small = small current large mag large current