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Q.

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats s−1  .The highest frequency is twice that of the lowest. Possible frequencies are

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a

120

b

48

c

40

d

80

answer is B, D.

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Detailed Solution

In n is frequency of first fork, then frequency of the last (10th fork) = n + 4(10 -1) = 2n
∴n = 36 and 2n = 72

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