Two wires are fixed in a sonometer. Their ten | vedclass

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(D) The frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}} = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}

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$10$

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(D) The frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}} = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 ho}} = \frac{1}{l} \sqrt{\frac{T}{\pi d^2 ho}}$,where $d$ is the diameter.Taking the ratio of frequencies for two wires:$\frac{n_1}{n_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2} 	imes \left(\frac{d_2}{d_1}ight)^2 	imes \frac{ho_2}{ho_1}}$Given ratios: $\frac{T_1}{T_2} = \frac{8}{1}$,$\frac{l_1}{l_2} = \frac{36}{35}$,$\frac{d_1}{d_2} = \frac{4}{1}$,$\frac{ho_1}{ho_2} = \frac{1}{2}$.Substituting these values:$\frac{n_1}{n_2} = \frac{35}{36} \sqrt{\frac{8}{1} 	imes \left(\frac{1}{4}ight)^2 	imes \frac{2}{1}}$$\frac{n_1}{n_2} = \frac{35}{36} \sqrt{8 	imes \frac{1}{16} 	imes 2} = \frac{35}{36} \sqrt{1} = \frac{35}{36}$.Since $n_1 = 360 \ Hz$ is the lower frequency,$n_1 = 360 \ Hz$ and $n_2 = 360 	imes \frac{36}{35} = 370 \ Hz$.Beat frequency = $|n_2 - n_1| = |370 - 360| = 10 \ Hz$.

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