Question

In physics lab, a student is performing experiment with resonance tube to find the speed of sound and radius of tube. For this, he used a resonance tube of length 120 cm. When the length of air column in tube is 16cm and 50cm, he obtained I and II resonance respectively, when a tuning fork of frequency 500 Hz is sounded just above the tube. Match the parameters in list-I with its suitable values in list-II

• A. A. Wavelength of sound (cm) | p
• B. B. height of liquid at II resonance (cm) | q
• C. C. Speed of sound (ms⁻¹) | r
• D. D. end correction (cm) | s
• E. E | minimum level of liquid at resonance (cm) | t
• F. (1) A→r, B→p, C→t, D → q, E→ s
• G. (2) A→s, B→t, C→r, D → q, E → p
• H. (3) A→t, B→s, C→r, D → p, E → q
• I. (4) A→s, B→t, C→r, D→ p, E → q

Answer 1

Answer: (I)

A→s, B→t, C→r, D→ p, E → q

Answer 2

The resonance tube acts as a closed organ pipe. The conditions for resonance are given by $l + e = (2n + 1)\frac{\lambda}{4}$. For I resonance ($l_1 = 16$ cm, $n=0$): $16 + e = \frac{\lambda}{4}$ For II resonance ($l_2 = 50$ cm, $n=1$): $50 + e = \frac{3\lambda}{4}$ Subtracting the first from the second: $34 = \frac{2\lambda}{4} = \frac{\lambda}{2}$, so $\lambda = 68$ cm. Substituting $\lambda$ into the first equation: $16 + e = \frac{68}{4} = 17$, so $e = 1$ cm. Speed of sound $v = f\lambda = 500 \text{ Hz} \times 68 \text{ cm} = 34000 \text{ cm/s} = 340 \text{ m/s}$. Height of liquid at II resonance = Total length - length of air column = $120 \text{ cm} - 50 \text{ cm} = 70 \text{ cm}$. The next resonance after $l_2=50$ cm ($n=1$) is for $n=2$: $l_3 + e = \frac{5\lambda}{4} \implies l_3 + 1 = \frac{5 \times 68}{4} = 85 \implies l_3 = 84$ cm. Water level = $120 - 84 = 36$ cm. The next resonance for $n=3$: $l_4 + e = \frac{7\lambda}{4} \implies l_4 + 1 = \frac{7 \times 68}{4} = 119 \implies l_4 = 118$ cm. Water level = $120 - 118 = 2$ cm. The minimum level of liquid at resonance is 2 cm. Thus, A→68 cm (s), B→70 cm (t), C→340 m/s (r), D→1 cm (p), E→2 cm (q).