Question

A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ______ cm.

(Take the speed of sound in air as 340 ms^-1.)

विकल्प

• 31

• 31

• 33

• 34

Answer 1

A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be 34 cm.

Explanation:

The resonant frequency of a closed organ pipe of length L is 

f = `(nν)/(4L)`

Here, n = odd positive Integer

ν = speed of sound in air

For L to be minimum, n = 1

∴ `250 = ν/(4L) ⇒ 250 = 340/(4L) ⇒ L = 34/(4 xx 25) = 0.34` m

⇒ L = 34 cm