Question:
In a resonance tube open at one end, the end correction is \(1.1\) cm. If the shortest length of resonating air column with a tuning fork is \(18\) cm, the next resonating length will be
In a resonance tube open at one end, the end correction is \(1.1\) cm. If the shortest length of resonating air column with a tuning fork is \(18\) cm, the next resonating length will be
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In a tube closed at one end, resonances occur at lengths with end correction equal to lambda/4, 3 lambda/4 and so on.
Updated On: Oct 1, 2026
- \(45.9\) cm
- \(49.6\) cm
- \(51.3\) cm
- \(56.2\) cm
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The Correct Option is D
Solution and Explanation
Step 1: Understand the concept
A resonance tube open at one end behaves as a closed pipe. Resonance occurs when \(L + e = \dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4}, \ldots\) where \(e\) is the end correction.
Step 2: First resonance
\(L_1 + e = \dfrac{\lambda}{4}\) gives \(18 + 1.1 = 19.1\) cm, so \(\lambda = 76.4\) cm.
Step 3: Next resonance
\[ L_2 + e = \frac{3\lambda}{4} = 3\times19.1 = 57.3\ \text{cm} \]
Step 4: Result
\(L_2 = 57.3 - 1.1 = 56.2\) cm, option (D). The value 51.3 cm would come from adding 33.3 cm, and 45.9 and 49.6 cm are not of the form \(3\times19.1 - 1.1\).
Final Answer:
The next resonating length is 56.2 cm. This is option (D). \[ \boxed{\text{(D) }56.2\ \text{cm}} \]
A resonance tube open at one end behaves as a closed pipe. Resonance occurs when \(L + e = \dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4}, \ldots\) where \(e\) is the end correction.
Step 2: First resonance
\(L_1 + e = \dfrac{\lambda}{4}\) gives \(18 + 1.1 = 19.1\) cm, so \(\lambda = 76.4\) cm.
Step 3: Next resonance
\[ L_2 + e = \frac{3\lambda}{4} = 3\times19.1 = 57.3\ \text{cm} \]
Step 4: Result
\(L_2 = 57.3 - 1.1 = 56.2\) cm, option (D). The value 51.3 cm would come from adding 33.3 cm, and 45.9 and 49.6 cm are not of the form \(3\times19.1 - 1.1\).
Final Answer:
The next resonating length is 56.2 cm. This is option (D). \[ \boxed{\text{(D) }56.2\ \text{cm}} \]
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