Q117 / 150 · MHT CET 2016 · Physics

If the end correction of an open pipe is $0.8 cm$ then the inner radius of that pipe will be

A
$\frac{1}{3} c m$
Correct answer
B
$\frac{2}{3} c m$
C
$\frac{3}{2} c m$
D
$0.2 c m$

Solution

For open organ pipe

∆ L = 1.2 \times r

∴ r = ∆ L \times \frac{1}{1.2} = \frac{0.8}{1.2} = \frac{2}{3} c m .

Hence, the correct answer is option: 'B'.

Relevant Questions

A tuning fork of frequency ' ' is held near the open end of a tube which is dipped in water and length of the tube is adjusted until resonan20 Apr, 2026 (M)Organ Pipe (Vibration of Air Column)
The third overtone of a closed pipe of length ' ' has the same frequency as the third overtone of the open pipe of length ' '. Both the pipe19 Apr, 2026 (M)Organ Pipe (Vibration of Air Column)
In a resonance tube open at one end, the end correction is cm. If the shortest length of resonating air column with a tuning fork is cm, the19 Apr, 2026 (E)Organ Pipe (Vibration of Air Column)
The lengths of the two pipes open at both ends are L and . If they are sounded together, the beat frequency will be ( = velocity of sound in18 Apr, 2026 (M)Organ Pipe (Vibration of Air Column)
Fifth overtone of an open pipe of length is in unison with the fifth overtone of the pipe closed at one end of length . The ratio to is18 Apr, 2026 (M)Organ Pipe (Vibration of Air Column)
The fundamental frequency ' ' of the tuning fork is 288 Hz, it will not resonate with the frequency18 Apr, 2026 (E)Organ Pipe (Vibration of Air Column)