QJEEM19Phe9e9: JEE Main 2019 Physics - Class 11 JEEM19Phe9e9 Solution: A wire of length 2L, is made by joining two wires A and B of…

Let mass per unit length of wires are $$\mu$$₁ and $$\mu$$₂ respectively

$$\because$$ Materials are same, so density $$\rho$$ will be 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 in both are same = T, let speed of wave in wires are V₁ and V₂

$$V_1=\frac{V_1}{2L}=\frac{V}{2L}\mathrm{\&}{V}_2=\frac{V_2}{2L}=\frac{V}{4L}$$

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 number of antinodes is 1 : 2.