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Question

A wire of density $$ 9 \times 10^3 Kg/m^3 $$ is stretched between two clamps 1 m apart and is stretched to an extension of $$ 4.9 \times 10^{-4} $$ meter, Young's modulus of material is $$ 9 \times 10^{10} N/m^2 $$. Then 


A
The lowest frequency of standing waveis 35 Hz
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B
The frequency of 1 st overtone is 70 Hz
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C
The frequency of 1 st overtone iis 105 Hz
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D
A and B both
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Solution

The correct option is A The lowest frequency of standing waveis 35 Hz
$$\begin{array}{l}\text { This is a fundamental standing wave } \\\qquad \begin{aligned}\Rightarrow L &=\frac{\lambda}{2}\\& \lambda=2 L \\\text { we know that, }&V=n\lambda\\\text { Also } & Y=\sqrt{\frac{T}{\mu}}\end{aligned}\end{array}$$

$$\begin{aligned}n_{0} &=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}\\\text { Given, } & \rho=a \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} \\& l=1 \mathrm{~m} \\& \Delta l=4.9\times10^{-4} \mathrm{~m} \\&Y=9\times10^{10}\mathrm{~N} / \mathrm{m}^{2}\end{aligned}$$

$$\begin{aligned}n_{0} &=\frac{1}{2 \times 1} \sqrt{\frac{V{A }\Delta l}{l A \rho}} \\&=\frac{1}{2}\sqrt{\frac{9\times10^{10} \times 4.9 \times 10^{-4}}{1\times 9 \times 10^{3}}}{10}\\n_{0}&=35\mathrm{~Hz}\end{aligned}$$

Physics

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