A) 4 : 9
B) 3 : 5
C) 1 : 4
D) 1 : 2
Correct Answer: D
Solution :
| (d) |
 |
| 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 |
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