NEET Solved Paper 2017 Question 25
Question: The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?
Options:
A) 40 Hz
B) 10 Hz
C) 20 Hz
D) 30 Hz
Show Answer
Answer:
Correct Answer: C
Solution:
Two successive frequencies of closed pipe $ \frac{nv}{4l}=220 $ ….(i)
$ \frac{(n+2)}{4l}=260 $ ….(ii)
Dividing (ii) by (i), we get $ \frac{n+2}{n}=\frac{260}{220}=\frac{13}{11} $ $ 11n+22=13n $
$ n=11. $ So, $ 11\frac{v}{4l}=220 $ $ \frac{v}{4l}=20 $ So fundamental frequency is 20 Hz.
BETA
