Two wires W_1 and W_2 have the same radius r | vedclass

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(B) For a sonometer wire,the frequency of vibration $n$ is given by $n = \frac{p}{2l} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinode

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$1:2$

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(B) For a sonometer wire,the frequency of vibration $n$ is given by $n = \frac{p}{2l} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$l$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Since $\mu = \pi r^2 \rho$,the frequency formula becomes $n = \frac{p}{2l} \sqrt{\frac{T}{\pi r^2 \rho}}$.For the composite wire,the frequency $n$ is the same for both parts $W_1$ and $W_2$. Also,$T$,$r$,and $l$ are the same for both wires.Thus,$n_1 = n_2 \implies \frac{p_1}{2l} \sqrt{\frac{T}{\pi r^2 \rho_1}} = \frac{p_2}{2l} \sqrt{\frac{T}{\pi r^2 \rho_2}}$.Simplifying,we get $\frac{p_1}{\sqrt{\rho_1}} = \frac{p_2}{\sqrt{\rho_2}}$.Given $\rho_2 = 4\rho_1$,we have $\frac{p_1}{\sqrt{\rho_1}} = \frac{p_2}{\sqrt{4\rho_1}} = \frac{p_2}{2\sqrt{\rho_1}}$.Therefore,$\frac{p_1}{p_2} = \frac{1}{2}$.

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