When an air column at 15 ^circ C and a tuning | vedclass

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Question

(A) Let $n$ be the frequency of the tuning fork and $f_{15}$ and $f_{10}$ be the frequencies of the air column at $15 ^\circ C$ and $10 ^\circ C$ resp

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(A) Let $n$ be the frequency of the tuning fork and $f_{15}$ and $f_{10}$ be the frequencies of the air column at $15 ^\circ C$ and $10 ^\circ C$ respectively.Given that at $15 ^\circ C$, $f_{15} - n = 4$, so $f_{15} = n + 4$.At $10 ^\circ C$, the beat frequency decreases by $1$, so $f_{10} - n = 3$, which means $f_{10} = n + 3$.The frequency of an air column is proportional to the speed of sound, $f \propto V$. Thus, $\frac{f_{15}}{f_{10}} = \frac{V_{15}}{V_{10}}$.Using $V_t = V_0 + 0.6t$, we have $V_{15} = 332 + 0.6(15) = 332 + 9 = 341, m/s$ and $V_{10} = 332 + 0.6(10) = 332 + 6 = 338, m/s$.Substituting these values: $\frac{n+4}{n+3} = \frac{341}{338}$.$338(n+4) = 341(n+3) \implies 338n + 1352 = 341n + 1023$.$3n = 329$. This calculation suggests a slight discrepancy in the provided options or parameters. Re-evaluating: If $f_{15} = \frac{V_{15}}{2l}$ and $f_{10} = \frac{V_{10}}{2l}$, then $\frac{f_{15}}{f_{10}} = \frac{341}{338} \approx 1.0088$. For $n=110$, $\frac{114}{113} \approx 1.0088$. Thus, $n = 110, Hz$ is the correct answer.

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