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Two strings $A$ and $B$ of lengths, $L_A=80 \mathrm{~cm}$ and $L_B=x \mathrm{~cm}$ respectively are used separately in a sonometer. The ratio of their densities $\left(d_A / d_B\right)$ is 0.81 . The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency the value of $x$ is :
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144
Given, $\frac{T_A}{T_B}=1, \quad \frac{L_A}{L_B}=\frac{80}{x}$
$\frac{D_A}{D_B}=\frac{2}{1}, \quad \frac{d_A}{d_B}=\frac{0.81}{1}$
Let $\mu_1$ and $\mu_2$ be the linear densities.
$\therefore \quad \frac{\mu_A}{\mu_B}=\left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B}=\left(\frac{2}{1}\right)^2 \times 0.81$
$=4 \times 0.81=3.24$
$\therefore \quad \frac{f_1}{f_2}=\frac{L_B}{L_A} \times \sqrt{\frac{T_A}{T_B} \times \frac{\mu_B}{\mu_{\mathrm{A}}}}$
$1=\frac{x}{80} \times \sqrt{1 \times \frac{1}{3.24}}$
$\therefore \quad x=80 \times \sqrt{3.24}$
or $\quad x=144$
$\frac{D_A}{D_B}=\frac{2}{1}, \quad \frac{d_A}{d_B}=\frac{0.81}{1}$
Let $\mu_1$ and $\mu_2$ be the linear densities.
$\therefore \quad \frac{\mu_A}{\mu_B}=\left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B}=\left(\frac{2}{1}\right)^2 \times 0.81$
$=4 \times 0.81=3.24$
$\therefore \quad \frac{f_1}{f_2}=\frac{L_B}{L_A} \times \sqrt{\frac{T_A}{T_B} \times \frac{\mu_B}{\mu_{\mathrm{A}}}}$
$1=\frac{x}{80} \times \sqrt{1 \times \frac{1}{3.24}}$
$\therefore \quad x=80 \times \sqrt{3.24}$
or $\quad x=144$
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