Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How to solve this problem- In Quick's Tube the sound intensity being detected changes from max to min for 2nd time, when the slidable tube is drawn apart by 9.0 cm. If speed of air 336 m/s, then frequency of their sounding source

In Quick's Tube the sound intensity being detected changes from max to min for 2nd time, when the slidable tube is drawn apart by 9.0 cm. If speed of air 336 m/s, then frequency of their sounding source

  • Option 1)

    2.8 kHz

  • Option 2)

    2.3 kHz

  • Option 3)

    2 kHz

  • Option 4)

    1 kHz

 

Answers (1)

best_answer

Total path diff=\frac{\lambda }{2}+\lambda =\frac{3\lambda }{2}

\frac{3\lambda }{2}=18cm

\lambda =12cm

f=\frac{v}{\lambda }=\frac{336}{12\times 10^{-2}}=2.8KHz

 

Quinck's tube -

This is an apparatus used to demonstrate the phenomena of interferance and also used to measure velocity of sound in air.

- wherein

x= \frac{\lambda }{2}

V=2\nu _{0}x

\nu _{0}= frequency

 

 


Option 1)

2.8 kHz

This is correct

Option 2)

2.3 kHz

This is incorrect

Option 3)

2 kHz

This is incorrect

Option 4)

1 kHz

This is incorrect

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams
anam-khan

JEE Main 2024 answer key out at jeemain.nta.ac.in; raise objections till April 14

Latest Question