A wire of density 8× 10^3 kg… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Waves & Oscillations

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A wire of density $8\times {10}^{3}\mathrm{kg}{m}^{-3}$ is stretched between two clamps $0.5m$ apart. The extension developed in the wire is $3.2\times {10}^{-4}m$ . If $Y=8\times {10}^{10}N{m}^{-2}$ , the fundamental frequency of vibration in the wire will be _____ $\mathrm{Hz}$

Official previous-year question

Held on 11 Apr 2023 · Verified 6 Jul 2026.

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Solution

Using the relation of Young's modulus,

$\frac{T}{A}=Y(\frac{\Delta L}{L})$

$\Rightarrow T=(\frac{Y\Delta L}{L}\times A)$

The linear mass density is $\mu =(\frac{m}{L})$ .

So,

$\frac{T}{\mu }=\frac{Y\Delta LA}{L(\frac{m}{L})}=\frac{Y(\Delta L)\times LA}{L(m)}=(\frac{Y\Delta L}{L})\times (\frac{1}{\rho })$

Substituting the values,

$\frac{T}{\mu }=\frac{8\times {10}^{10}\times 3.2\times {10}^{-4}}{0.5}\times (\frac{1}{8\times {10}^{3}})=6.4\times {10}^{3}$

$\Rightarrow \frac{T}{\mu }=64\times {10}^{2}$

The fundamental frequency is given by $f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ .

$\Rightarrow \sqrt{\frac{T}{\mu }}=8\times 10=80m{s}^{-1}$

Therefore,

$f=(\frac{80}{1})=80\mathrm{Hz}$

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Solution

Using the relation of Young's modulus,

$\frac{T}{A}=Y(\frac{\Delta L}{L})$

$\Rightarrow T=(\frac{Y\Delta L}{L}\times A)$

The linear mass density is $\mu =(\frac{m}{L})$ .

So,

$\frac{T}{\mu }=\frac{Y\Delta LA}{L(\frac{m}{L})}=\frac{Y(\Delta L)\times LA}{L(m)}=(\frac{Y\Delta L}{L})\times (\frac{1}{\rho })$

Substituting the values,

$\frac{T}{\mu }=\frac{8\times {10}^{10}\times 3.2\times {10}^{-4}}{0.5}\times (\frac{1}{8\times {10}^{3}})=6.4\times {10}^{3}$

$\Rightarrow \frac{T}{\mu }=64\times {10}^{2}$

The fundamental frequency is given by $f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ .

$\Rightarrow \sqrt{\frac{T}{\mu }}=8\times 10=80m{s}^{-1}$

Therefore,

$f=(\frac{80}{1})=80\mathrm{Hz}$

Did you get this right?

Solution

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Did you get this right?

Solution

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Other Physics topics for JEE Main

Mechanics

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