Question

One end of two wires of the same metal and of the same length (with radius r and 2r) are joined together. The wire is used as sonometer wire and the junction is placed in between two bridges. Tension T is applied to the wire, if at a junction a node is formed, then the ratio of number of loops formed in the wires will be

• A. 1 : 2
• B. 2 : 3
• C. 3 : 4
• D. 4 : 5

Answer 1

The correct option is A 1 : 2

From the formula

$F=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where $n=n^m$ harmonic frequency

$=$ no of loops

$n\alpha \sqrt{\mu }$

Now, $\mu =\frac{m}{L}=\frac{\rho V}{L}=\frac{\rho \times \pi r^{2}L}{L}$

$\Rightarrow n\alpha r$ ($r$ = radius of wire)

$\therefore$ Ratio of no. loops formed in the wire under same tension and length.

$\frac{n_1}{n_2}=\frac{r}{2r}=\frac{1}{2}$

$[1:2ratio]$ Ans

$\therefore$ Option A is correct.