Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

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

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

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.
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
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: