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A wire of length $2\,L$ is made by joining two wires $A$ and $B$ of same lengths but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$ then the ratio $p : q$ is
$1 : 4$
$1 : 2$
$3 : 5$
$4 : 9$
Solution
Let mass per unit length of wires are $\mu_{1}$ and $\mu_{2}$ respectively.
$\because$ Materials are same, so density $\rho$ is same.
$\mu_{1}=\frac{\rho \pi r^{2} L}{L}=\mu \text { and } \mu_{2}=\frac{\rho 4 \pi r^{2} L}{L}=4 \mu$
Tension in both are same $=\mathrm{T},$ let speed of wave in wires are $V_{1}$ and $V_{2}$
$V_{1}=\frac{V_{1}}{2 L}=\frac{V}{2 L} \& f_{02}=\frac{V_{2}}{2 L}=\frac{V}{4 L}$
Frequency at which both resonate is $L.C.M.$ of both frequencies i.e. $\frac{\mathrm{V}}{2 \mathrm{L}}$
Hence number of loops in wires are $1$ and $2$ respectively
So, ratio of number of antinodes is $1:2$
