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Two wires $W_1$ and $W_2$ have the same radius $r$ and respective densities ${\rho _1}$ and ${\rho _2}$ such that ${\rho _2} = 4{\rho _1}$. 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 an tin odes formed in $W_1$ to $W_2$ is
$1:1$
$1 : 2$
$1 : 3$
$4 : 1$
Solution
$\begin{array}{l}
{n_1} = {n_2}\\
T \to Same\\
r \to Same\\
l \to Same
\end{array}$
Frequency of vibration
$\mathrm{n}=\frac{\mathrm{p}}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}^{2} \rho}}$
As $\mathrm{T}, \mathrm{r},$ and $l$ are same for both the wires
$n_{1}=n_{2}$
$\frac{p_{1}}{\sqrt{\rho_{1}}}=\frac{p_{2}}{\sqrt{p_{2}}}$
$\frac{p_{1}}{p_{2}}=\frac{1}{2} \quad \because \rho_{2}=4 \rho_{1}$
