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Two wires are fixed on a sonometer. Their tensions are in the ratio 8 : 1, the lengths are in the ratio 26 : 25, the diameters are in the ratio 4 : 1 and the densities are in the ratio of 1 : 2. If the frequency of higher pitch is 260Hz, the number of beats produced per second is

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answer is B.

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Detailed Solution

$v = \frac{1}{D l} \sqrt{\frac{T}{π d}}$

$\frac{v_{1}}{v_{2}} = \frac{D_{2}}{D_{1}} \frac{l_{2}}{l_{1}} \sqrt{\frac{T_{1}}{T_{2}} \times \frac{d_{2}}{d_{1}}}$

$\frac{v_{1}}{v_{2}} = \frac{1}{4} \frac{25}{26} \sqrt{\frac{8}{1} \times \frac{2}{1}}$

$\frac{v_{1}}{v_{2}} = \frac{25}{26}$

$\Rightarrow$ Second wire is of higher pitch. Its frequency is 260Hz

$\frac{v_{1}}{260} = \frac{25}{26}$

$v_{1} = 250 Hz$

Number of beats per second - $v_{2} - v_{1} = 10$

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