The fundamental frequency of vibration of a stretched string is given by:

$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $

where

$L$ = length of the string,
$T$ = tension,
$\mu$ = mass per unit length.

Step 1: Find $\mu$ in terms of density and diameter

The mass per unit length, $\mu = \text{density} \times \text{area of cross-section}$.

For a wire of diameter $d$:

$ \text{Area} = \frac{\pi d^2}{4} $

So,

$ \mu = e \times \frac{\pi d^2}{4} $

Step 2: Write expressions for $n_A$ and $n_B$

For wire A:

Length = $L$
Tension = $T$
Density = $e$
Diameter = $d$

So,

$ \mu_A = e \frac{\pi d^2}{4} $

$ n_A = \frac{1}{2L} \sqrt{\frac{T}{e \frac{\pi d^2}{4}}} $

For wire B:

Length = $2L$
Tension = $2T$
Density = $2e$
Diameter = $3d$

So,

$ \mu_B = 2e \times \frac{\pi (3d)^2}{4} = 2e \times \frac{\pi \times 9d^2}{4} = \frac{18e\pi d^2}{4} $

$ n_B = \frac{1}{2(2L)} \sqrt{\frac{2T}{\frac{18e\pi d^2}{4}}} $

Step 3: Simplify $n_A$ and $n_B$

$n_A$:

$ n_A = \frac{1}{2L} \sqrt{\frac{T}{e \frac{\pi d^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{e\pi d^2}} $

$n_B$:

$ n_B = \frac{1}{4L} \sqrt{\frac{2T}{\frac{18e\pi d^2}{4}}} = \frac{1}{4L} \sqrt{\frac{2T \times 4}{18e\pi d^2}} = \frac{1}{4L} \sqrt{\frac{8T}{18e\pi d^2}} = \frac{1}{4L} \sqrt{\frac{4T}{9e\pi d^2}} $

Step 4: Find the ratio $\frac{n_B}{n_A}$

Take the simplified forms:

$ n_A = \frac{1}{2L} \sqrt{\frac{4T}{e\pi d^2}} $

$ n_B = \frac{1}{4L} \sqrt{\frac{4T}{9e\pi d^2}} $

So,

$ \frac{n_B}{n_A} = \frac{\frac{1}{4L} \sqrt{\frac{4T}{9e\pi d^2}}}{\frac{1}{2L} \sqrt{\frac{4T}{e\pi d^2}}} $

$ = \frac{1}{4L} \div \frac{1}{2L} \times \frac{\sqrt{\frac{4T}{9e\pi d^2}}}{\sqrt{\frac{4T}{e\pi d^2}}} $

$ = \frac{1}{2} \times \frac{\sqrt{\frac{4T}{9e\pi d^2}}}{\sqrt{\frac{4T}{e\pi d^2}}} $

Let's write the square roots explicitly:

$ \frac{\sqrt{\frac{4T}{9e\pi d^2}}}{\sqrt{\frac{4T}{e\pi d^2}}} = \sqrt{\frac{\frac{4T}{9e\pi d^2}}{\frac{4T}{e\pi d^2}}} = \sqrt{\frac{4T}{9e\pi d^2} \times \frac{e\pi d^2}{4T}} = \sqrt{\frac{1}{9}} = \frac{1}{3} $

So:

$ \frac{n_B}{n_A} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} $

Final Ratio:

$ n_B : n_A = 1 : 6 $

Correct option: Option C

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