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(D) Let the frequency of the first tuning fork be $n_1 = n$.Since there are $10$ tuning forks and each pair of adjacent forks produces $4 	ext{ beats

Vedclass · Accepted Answer

$72 	ext{ Hz}$ and $36 	ext{ Hz}$

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(D) Let the frequency of the first tuning fork be $n_1 = n$.Since there are $10$ tuning forks and each pair of adjacent forks produces $4 	ext{ beats/sec}$,the frequencies form an arithmetic progression with common difference $d = 4 	ext{ Hz}$.The frequency of the $N$-th tuning fork is given by $n_N = n_1 + (N - 1)d$.For $N = 10$ and $d = 4$,the highest frequency is $n_{10} = n + (10 - 1) 	imes 4 = n + 36$.According to the problem,the highest frequency is twice the lowest frequency,so $n_{10} = 2n$.Equating the two expressions for $n_{10}$:$2n = n + 36$$n = 36 	ext{ Hz}$.Thus,the lowest frequency is $n_1 = 36 	ext{ Hz}$ and the highest frequency is $n_{10} = 2 	imes 36 = 72 	ext{ Hz}$.

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce $4 \text{ beats/sec}$. The highest frequency is twice the lowest. The possible highest and lowest frequencies are:

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