A wire of length 30.0 cm and mass 10.0 g is secured at both ends...

The fundamental frequency of the closed organ pipe is given by:

$f = \dfrac{v}{4L_c}$

Substituting $v = 340 \text{ m s}^{-1}$ and $L_c = 0.5 \text{ m}$:

$f = \dfrac{340}{4 \times 0.5} = 170 \text{ Hz}$

Since the wire is in resonance with the pipe, its fundamental frequency is also $170 \text{ Hz}$. The fundamental frequency of a stretched wire is given by:

$f = \dfrac{1}{2L_w} \sqrt{\dfrac{T}{\mu}}$

Here, the length of the wire is $L_w = 0.3 \text{ m}$ and its linear mass density is $\mu = \dfrac{m}{L_w} = \dfrac{10 \times 10^{-3}}{0.3} = \dfrac{1}{30} \text{ kg m}^{-1}$.

Substituting the known values into the frequency formula:

$170 = \dfrac{1}{2 \times 0.3} \sqrt{\dfrac{T}{1/30}}$

$170 \times 0.6 = \sqrt{30T}$

$102 = \sqrt{30T}$

Squaring both sides:

$10404 = 30T$

$T = \dfrac{10404}{30} = 346.8 \text{ N}$

Rounding off, the tension in the wire is approximately $347 \text{ N}$.