1. Fundamental frequencies of the pipes

For a closed organ pipe of length $L_c$, fundamental frequency is $f_c = \frac{v}{4L_c}$

For an open organ pipe of length $L_o$, fundamental frequency is $f_o = \frac{v}{2L_o}$

Given: $L_c = 150\text{ cm} = 1.5\text{ m}, \qquad L_o = 350\text{ cm} = 3.5\text{ m}$

So, $f_c = \frac{v}{4(1.5)} = \frac{v}{6}$ $f_o = \frac{v}{2(3.5)} = \frac{v}{7}$

2. Using beat frequency

Beat frequency is the absolute difference of the two frequencies: $|f_c - f_o| = 7$

Substitute the expressions: $\left|\frac{v}{6} - \frac{v}{7}\right| = 7$

$v\left|\frac{1}{6} - \frac{1}{7}\right| = 7$

$v\left(\frac{1}{42}\right) = 7$

$v = 7 \times 42 = 294\text{ m/s}$

3. Final answer

$\boxed{294}$

4. Comparison with stored answer

Stored correct answer = $294$

My derived answer matches the stored answer.