A wire of length 2L, is made by joining two wires A and B of same length 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

Let mass per unit length of wires are ${\mu }_1$ and ${\mu }_2$ respectively. Materials are same,  so density $\rho$ is same.
$\therefore {\mu }_1=\frac{\rho \pi r^{2}L}{L}=\mu$  and  ${\mu }_2=\frac{\rho 4\pi r^{2}L}{L}=4\mu$$$
Tension is both are same = T, let speed of wave in wires are $v_1$ and  $v_2$
 $v_1=\sqrt{\frac{T}{\mu }}=v;v_2=\sqrt{\frac{T}{4\mu }}=\frac{v}{2}$
So fundamental frequencies in both wires are  $f{o}_1=\frac{V_1}{2L}=\frac{V}{2L}\&f{o}_2=\frac{V_2}{2L}=\frac{V}{4L}$
Frequency at which both resonate is L.C.M of both frequencies  $i.e.,\frac{V}{2L}$
Hence, number of loops in wires are 1 and 2 respectively. 
So, ratio of no.of antinodes is 1 : 2