Q128 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
Resonance is produced between a turning fork and a resonance column tube with upper end open and lower end closed by water surface. If the frequency of the tuning fork is 800Hz, and first two resonances are observed at lengths 9.75 cm, 31.25 cm, Find the length at which the third resonance occurs? Also find the speed of sound.
- Correct answerA$52.75$ cm and $344\ ms^{-1}$
- B$43.10$ cm and $340\ ms^{-1}$
- C$62.60$ cm and $335\ ms^{-1}$
- D$31.25$ cm and $330\ ms^{-1}$
Solution
Let $L_1$, $L_2$, and $L_3$ be the lengths of the first, second, and third resonances respectively, and $e$ be the end correction.
The resonance lengths for a tube closed at one end are given by:
$L_1 + e = \dfrac{\lambda}{4}$
$L_2 + e = \dfrac{3\lambda}{4}$
$L_3 + e = \dfrac{5\lambda}{4}$
Subtracting the first equation from the second:
$L_2 - L_1 = \dfrac{\lambda}{2}$
Substituting the given values:
$31.25 - 9.75 = \dfrac{\lambda}{2}$
$21.50 = \dfrac{\lambda}{2} \Rightarrow \lambda = 43.0\text{ cm} = 0.43\text{ m}$
The speed of sound is:
$v = f \lambda = 800 \times 0.43 = 344\text{ ms}^{-1}$
For the third resonance length:
$L_3 - L_2 = \dfrac{\lambda}{2}$
$L_3 = 31.25 + 21.50 = 52.75\text{ cm}$
Answer: $52.75$ cm and $344\ ms^{-1}$
The resonance lengths for a tube closed at one end are given by:
$L_1 + e = \dfrac{\lambda}{4}$
$L_2 + e = \dfrac{3\lambda}{4}$
$L_3 + e = \dfrac{5\lambda}{4}$
Subtracting the first equation from the second:
$L_2 - L_1 = \dfrac{\lambda}{2}$
Substituting the given values:
$31.25 - 9.75 = \dfrac{\lambda}{2}$
$21.50 = \dfrac{\lambda}{2} \Rightarrow \lambda = 43.0\text{ cm} = 0.43\text{ m}$
The speed of sound is:
$v = f \lambda = 800 \times 0.43 = 344\text{ ms}^{-1}$
For the third resonance length:
$L_3 - L_2 = \dfrac{\lambda}{2}$
$L_3 = 31.25 + 21.50 = 52.75\text{ cm}$
Answer: $52.75$ cm and $344\ ms^{-1}$
Hence, the correct answer is option: 'A'.