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(A) Let the frequency of the first tuning fork be $n 	ext{ Hz}$.Since the forks are arranged in an increasing order of frequencies with a beat freque

Vedclass · Accepted Answer

$220$

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(A) Let the frequency of the first tuning fork be $n 	ext{ Hz}$.Since the forks are arranged in an increasing order of frequencies with a beat frequency of $4 	ext{ Hz}$ between consecutive forks,the frequencies form an arithmetic progression $(AP)$.The number of tuning forks is $N = 56$.The common difference $d = 4 	ext{ Hz}$.The frequency of the $N^{th}$ fork is given by $f_N = n + (N - 1)d$.Substituting the values,$f_{56} = n + (56 - 1) 	imes 4 = n + 55 	imes 4 = n + 220$.According to the problem,the frequency of the last fork is the octave of the first,which means $f_{56} = 2n$.Equating the two expressions: $n + 220 = 2n$.Solving for $n$: $n = 220 	ext{ Hz}$.

$56$ tuning forks are arranged in increasing order of frequencies in a series such that each fork gives $4 \text{ beats per second}$ with the previous one. The frequency of the last fork is the octave of the first. The frequency of the first fork is ..... $Hz$.

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