A wire is stretched between two rigid supports. It is observed that wire resonates at the frequencies f1, f2, f3 and f4(f4>f3>f2>f1) forming 2,3,4 and 5 loops respectively. The ratio of any two resonance frequencies will be minimum for difference in the loops to be:
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
1
b
2
c
3
d
zero
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Let f1 = mv2L = 2v2L. Similarly f4 = (n+3)2L = 5v2Lf = f1f4 will be minimum f = f1f4 = nn+3 f will be minimum for difference in loops = 3.
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
A wire is stretched between two rigid supports. It is observed that wire resonates at the frequencies f1, f2, f3 and f4(f4>f3>f2>f1) forming 2,3,4 and 5 loops respectively. The ratio of any two resonance frequencies will be minimum for difference in the loops to be: