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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2021 · 31 August · Morning Shift · Physics · Waves and Sound

A wire having a linear mass density 9 . 0 × 10 ^ - 4 kg m ^ - 1 is...

A wire having a linear mass density 9.0×10−4kgm−19 . 0 \times {10}^{- 4}kg{m}^{- 1} is stretched between two rigid supports with a tension of 900N.900N . The wire resonates at a frequency of 500Hz.500Hz . The next higher frequency at which the same wire resonates is 550Hz.550Hz . The length of the wire is ___________ m.m .

Correct answer

0

Step-by-step solution

fn=nv2ℓ=500{f}_{n} = \dfrac{nv}{2ℓ} = 500
fn+1=(n+1)v2ℓ=550{f}_{n + 1} = \dfrac{( n + 1 ) v}{2ℓ} = 550
n+1n=1110\dfrac{n + 1}{n} = \dfrac{11}{10}
Thus, n=10n = 10
Thus, ℓ=nv2fn(v=Tμ)ℓ = \dfrac{nv}{2{f}_{n}}\left(v = \sqrt{\dfrac{T}{\mu}}\right)
=102×500×9009×10−4=10m= \dfrac{10}{2 \times 500} \times \sqrt{\dfrac{900}{9 \times {10}^{- 4}}} = 10m
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