Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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 speed of sound in air as 340 m s-1 )
PhysicsWaves and SoundJEE MainJEE Main 2021 (27 Aug Shift 2)
Solution:
1729 Upvotes Verified Answer
The correct answer is: 34

For closed organ pipe

f=(2n+1)v4l

for minimum length, n=0

f0=v4l

l=v4f0

=3404×250=34 cm

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.