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Two wires W₁ and W₂ have the same radius r and respective densities $ρ_{1} a n d ρ_{2}$ such that $ρ_{1} = 16 ρ_{2}$ . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary waves is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W₁ and W₂ is:

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answer is 4.

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Detailed Solution

$n_{1} = n_{2} T \to S a m e r \to S a m e l \to S a m e$

Frequency of vibration
$n = \frac{p}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$
As T, r, and l are same for both the wires
$n_{1} = n_{2} ρ_{1} = 16 ρ_{2} \frac{P_{1}}{\sqrt{ρ_{1}}} = \frac{P_{2}}{\sqrt{ρ_{2}}} \Rightarrow \frac{ρ_{1}}{ρ_{2}} = 4$

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