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Two uniform wire are vibrating simultaneously in their fundamental modes. The tensions, lengths, diameters, and the densities of the two wires are in the ratios, 8 : 1 ; 36 : 35 ; 4 : 1 ; 1 : 2 respectively. If the note of higher pitch has a frequency of 360Hz the number of beats produced per second is

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answer is B.

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Detailed Solution

$n = \frac{1}{2 L} \sqrt{\frac{T}{μ}}$ ; $μ = (\frac{{πd}^{2}}{4}) ρ$

$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{L_{2}}{L_{1}} \frac{d_{2}}{d_{1}} \sqrt{\frac{T_{1}}{T_{2}} \cdot \frac{ρ_{2}}{ρ_{1}}}$

$\frac{n_{1}}{n_{2}} = \frac{35}{36} \times \frac{1}{4} \sqrt{\frac{8}{1} \times \frac{2}{1}} = \frac{35}{36}$

$\Rightarrow n_{1} < n_{2} \Rightarrow \frac{n_{1}}{360} < \frac{35}{36} \Rightarrow n_{1} = 350 Hz$
$\Rightarrow n_{2} - n_{1} = 360 - 350 = 10$

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