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Q.

Fifty six tuning forks are arranged in order of increasing frequencies so that each fork gives 4 beats per seconds with the next one. The last fork gives the octave of the first. The frequency of the first is n×110 Hz. The value of n is

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answer is 2.

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Detailed Solution

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tn=a+(n1)d[A.P]2n=n+(561)4n=55×4n=220Hz

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Fifty six tuning forks are arranged in order of increasing frequencies so that each fork gives 4 beats per seconds with the next one. The last fork gives the octave of the first. The frequency of the first is n×110 Hz. The value of n is