A uniform tube of length $$60.5$$ cm is held vertically with its lower end dipped in water. A sound source of frequency $$500$$ Hz sends sound waves into the tube. When the length of tube above water is $$16$$ cm and again when it is $$50$$ cm, the tube resonates with the source of sound. Two lowest frequencies (in Hz), to which tube will resonate when it is taken out of water, are (approximately).

The air column behaves as a pipe closed at the water-surface end and open at the top end. For a pipe closed at one end, the resonance (standing-wave) condition is

$$L_{\text{eff}} = \frac{(2n-1)\,\lambda}{4},\qquad n = 1,2,3,\dots$$

where $$L_{\text{eff}}$$ is the effective length of the air column and $$\lambda$$ the wavelength of sound. Because the open end is not a sharp boundary, the antinode actually forms a little outside the tube. That shift is the end-correction $$e\;(\approx 0.6\,r)$$, so

$$L_{\text{eff}} = L + e$$

where $$L$$ is the measured (geometrical) length of air in the tube.

Given two successive resonances at $$L_1 = 16\text{ cm}=0.16\text{ m},\qquad L_2 = 50\text{ cm}=0.50\text{ m},$$ the difference between successive resonant lengths of a closed pipe is

$$L_2-L_1 = \frac{\lambda}{2}\; \Longrightarrow \; \lambda = 2(L_2-L_1) = 2(0.50-0.16)=0.68\text{ m}.$

The speed of sound in air is therefore

$$v = f\,$$\lambda$$ = 500\,$$\text{Hz}$$$$\times$$0.68\,$$\text{m}$$=340\;$$\text{m s}^{-1}$$.$$

Using the first resonance ($$n=1$$) to obtain the end correction:

$$L_{$$\text{eff}$$} = $$\frac{\lambda}{4} = \frac{0.68}{4}$$=0.17$$\text{ m}$$,$$ $$e = L_{$$\text{eff}$$}-L_1 = 0.17-0.16 = 0.01$$\text{ m}$$ = 1$$\text{ cm}$$.$$

When the tube is taken out of water, both ends are open. For a pipe open at both ends, the standing-wave condition is

$$L_{$$\text{eff(open)}$$} = $$\frac{n\,\lambda}{2}$$,\qquad n = 1,2,3,$$\dot$$s$$

Now the antinode shift occurs at both ends, so

$$L_{$$\text{eff(open)}$$} = L_0 + 2e,$$

where $$L_0 = 60.5$$\text{ cm}$$=0.605$$\text{ m}$$$$ is the actual length of the tube in air.

Hence $$L_{$$\text{eff(open)}$$} = 0.605 + 2(0.01) = 0.625$$\text{ m}$$.$$

The fundamental frequency ($$n=1$$) for the open pipe is

$$f_1 = $$\frac{v}{2L_{\text{eff(open)}$$}} = $$\frac{340}{2\times0.625} = \frac{340}{1.25}$$$$\approx$$272$$\text{ Hz}$$.$$

The next higher resonant frequency ($$n=2$$) is simply double the fundamental:

$$f_2 = 2f_1 $$\approx$$ 2$$\times$$272 = 544$$\text{ Hz}$$.$$

Therefore, the two lowest frequencies to which the tube will resonate when it is completely in air are approximately $$272$$\text{ Hz}$$$$ and $$544$$\text{ Hz}$$$$.

Option D which is: $$272, 544$$