An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed

An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive nodes are formed at distances 16 cm and 46 cm from the open end. The speed of sound in air in the pipe is

Options

(a) 230 m/s
(b) 300 m/s
(c) 320 m/s
(d) 360 m/s

Correct Answer:

300 m/s

Explanation:

Two successive modes are separated by λ/2. λ/2 = (46-16)=30 λ=60 cm=0.6 m v=nλ=500×0.6=300 m/s.

admin:

View Comments (2)

Related Questions

  1. A parallel plate air capacitor has capacity C, distance of seperation between plate
  2. In the half wave rectifier circuit operating from 50 Hz mains frequency,
  3. Thin uniform rod of length L, cross-sectional area A and density ρ is rotated
  4. Which of the following atoms has the lowest ionisation potential?
  5. The ratio of specific charge of an α-particle to that of a proton is