Two open organ pipes of lengths $$60 \text{ cm}$$ and $$90 \text{ cm}$$ resonate at $$6^{th}$$ and $$5^{th}$$ harmonics respectively. The difference of frequencies for the given modes is ___________ Hz. (Velocity of sound in air $$= 333 \text{ m/s}$$)

We need to find the difference in frequencies between the 6th harmonic of a 60 cm open pipe and the 5th harmonic of a 90 cm open pipe. Recall that for an open organ pipe, the frequency of the $$n$$th harmonic is given by $$f_n = \frac{nv}{2L}$$ where $$v$$ is the speed of sound, $$L$$ is the length of the pipe, and $$n = 1, 2, 3, \ldots$$.

First, applying this formula to the 6th harmonic of the first pipe, with $$L_1$$ = 60 cm = 0.6 m, we obtain $$f_1 = \frac{6 \times 333}{2 \times 0.6} = \frac{1998}{1.2} = 1665 \, \text{Hz}$$.

Next, for the 5th harmonic of the second pipe, with $$L_2$$ = 90 cm = 0.9 m, we find $$f_2 = \frac{5 \times 333}{2 \times 0.9} = \frac{1665}{1.8} = 925 \, \text{Hz}$$.

Subtracting these values gives $$\Delta f = f_1 - f_2 = 1665 - 925 = 740 \, \text{Hz}$$.

The correct answer is 740 Hz.