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(C) Let the frequency of the first tuning fork be $n_1 = n \, 	ext{Hz}$.Since there are $N = 50$ tuning forks and each gives $4 \, 	ext{beats/sec}$

Vedclass · Accepted Answer

$196$

Vedclass · Answer

(C) Let the frequency of the first tuning fork be $n_1 = n \, 	ext{Hz}$.Since there are $N = 50$ tuning forks and each gives $4 \, 	ext{beats/sec}$ with the previous one, the frequencies form an arithmetic progression with common difference $d = 4 \, 	ext{Hz}$.The frequency of the $N$-th tuning fork is given by $n_N = n_1 + (N - 1)d$.Substituting the values, $n_{50} = n + (50 - 1) 	imes 4 = n + 49 	imes 4 = n + 196$.Given that the frequency of the last fork is the octave of the first, we have $n_{50} = 2n$.Equating the two expressions: $2n = n + 196$.Solving for $n$: $n = 196 \, 	ext{Hz}$.

$50$ tuning forks are arranged in increasing order of their frequencies such that each gives $4 \, \text{beats/sec}$ with its previous tuning fork. If the frequency of the last fork is the octave of the first, then the frequency of the first tuning fork is ... $Hz$.

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