Waves

JEE Main level, numerical answer, medium

68 of 142

Question 68 of 142

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.

Show answer and solution

Answer: 10

fn=nv2ℓ=500{f}_{n}=\dfrac{nv}{2\ell }=500
fn+1=(n+1)v2ℓ=550{f}_{n+1}=\dfrac{(n+1)v}{2\ell }=550
n+1n=1110\dfrac{n+1}{n}=\dfrac{11}{10}
Thus, n=10n=10
Thus, ℓ=nv2fn(v=Tμ)\ell =\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

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Standing waves on a string simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Waves (Vibration of String), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.