Q125 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
A tuning fork of frequency '$n$' is held near the open end of a tube which is dipped in water and length of the tube is adjusted until resonance occurs. If the two shortest lengths that produce resonance are $l_1$ and $l_2$, the speed of sound in air is (neglect end correction)
- Correct answerA$2n(l_2 - l_1)$
- B$n(l_2 - l_1)$
- C$\dfrac{n}{2}(l_2 - l_1)$
- D$\dfrac{2n}{(l_2 - l_1)}$
Solution
For a resonance tube, the water level acts as a closed end, forming a closed organ pipe.
The first resonance occurs at length $l_1 = \dfrac{\lambda}{4}$
The second resonance occurs at length $l_2 = \dfrac{3\lambda}{4}$
The difference between the two resonating lengths is:
$l_2 - l_1 = \dfrac{3\lambda}{4} - \dfrac{\lambda}{4} = \dfrac{\lambda}{2}$
$\Rightarrow \lambda = 2(l_2 - l_1)$
The speed of sound is given by $v = n\lambda$
Substituting the value of $\lambda$, we get:
$v = 2n(l_2 - l_1)$
Answer: $2n(l_2 - l_1)$
The first resonance occurs at length $l_1 = \dfrac{\lambda}{4}$
The second resonance occurs at length $l_2 = \dfrac{3\lambda}{4}$
The difference between the two resonating lengths is:
$l_2 - l_1 = \dfrac{3\lambda}{4} - \dfrac{\lambda}{4} = \dfrac{\lambda}{2}$
$\Rightarrow \lambda = 2(l_2 - l_1)$
The speed of sound is given by $v = n\lambda$
Substituting the value of $\lambda$, we get:
$v = 2n(l_2 - l_1)$
Answer: $2n(l_2 - l_1)$
Hence, the correct answer is option: 'A'.