Questions
When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is
detailed solution
Correct option is D
nA = Known frequency = 341 Hz, nB = ?x = 6 bps, which is decreasing (i.e. x↓) after loading (from 6 to 1 bps)Unknown tuning fork is loaded so nB↓Hence nA – nB↓ = x↓ ... (i) → Wrong nB↓ – nA = x↓ ... (ii) → Correct⇒ nB = nA + x = 341 + 6 = 347 Hz.Talk to our academic expert!
Similar Questions
A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance 50 cm the tension in the wire is 100 N the wire vibrating with its fundamental frequency and vibrating tuning fork together produces 5 beats per second. The tension in the wire is then reduced to 81 N, when the two are excited, beats are heard at the same rate. Calculate frequency of the fork.
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests