A) 4 : 9
B) 3 : 5
C) 1 : 4
D) 1 : 2
Correct Answer: D
Solution :
Let mass per unit length of wires are\[{{\mu }_{1}}\]and \[{{\mu }_{2}}\]respectively. \[\because \]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 in 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 \[{{\text{f}}_{01}}=\frac{{{V}_{1}}}{2L}=\frac{V}{2L}\And {{\text{f}}_{02}}=\frac{{{V}_{2}}}{2L}=\frac{V}{4L}\] Frequency at which both resonate is L.C.M of both frequencies i.e.\[\frac{V}{2L}.\] Hence no. of loops in wires are 1 and 2 respectively. So, ratio of no. of antinodes is 1 : 2.You need to login to perform this action.
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