The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is 5/x. The value of x is ______.

We need to find x where the ratio of lengths of closed pipe to open pipe is 5/x.

Key formulas:

For a closed organ pipe, the nth harmonic frequency (only odd harmonics): $$f_n = \frac{nv}{4L_c}$$ where n = 1, 3, 5, ...

For an open organ pipe, the nth harmonic: $$f_n = \frac{nv}{2L_o}$$ where n = 1, 2, 3, ...

Fifth harmonic of closed pipe = First harmonic of open pipe.

Fifth harmonic of closed pipe: $$f = \frac{5v}{4L_c}$$

First harmonic of open pipe: $$f = \frac{v}{2L_o}$$

Setting them equal:

$$\frac{5v}{4L_c} = \frac{v}{2L_o}$$

$$\frac{5}{4L_c} = \frac{1}{2L_o}$$

$$\frac{L_c}{L_o} = \frac{5 \times 2}{4} = \frac{10}{4} = \frac{5}{2}$$

So $$\frac{L_c}{L_o} = \frac{5}{2}$$, which means x = 2.

Therefore, the value of x is Option 3: 2.