Answer: (2) 1:3
- Frequency of A = n1
- Frequency of B = n2
∴ 2n1 = 3n2 or
\(\begin{array}{l}\frac{2}{2l_{1}}\sqrt{\frac{T}{m_{1}}} = \frac{3}{2l_{2}}\sqrt{\frac{T}{m_{2}}}\end{array} \)
.
Where m = mass/length
m = (cross sectional area × l × ρ)/l
m = cross sectional area × ρ
\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{2}{3}\sqrt{\frac{m_{2}}{m_{1}}}= \sqrt{\frac{a_{2}\rho}{a_{1}\rho}}\end{array} \)
Where
- a = Cross-sectional area(a = πr2)
- ρ = Density of the material
Since density is same, above equation becomes
\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{2}{3}\sqrt{\frac{r_{2}^{2}}{r_{1}^{2}}}\end{array} \)
.
\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{2}{3}\sqrt({\frac{1}{2})^{2}}\end{array} \)
.
\(\begin{array}{l}\frac{l_{1}}{l_{2}}= \frac{1}{3}\end{array} \)
