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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/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

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a

80 and 40

b

100 and 50

c

44 and 22

d

72 and 36

answer is D.

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Detailed Solution

Using nLast = nFirst + (N – 1)x where N = Number of tuning fork in series         x = beat frequency between two successive forks ⇒ 2n = n + (10 – 1) ´ 4 ⇒ n = 36 Hz ∴ nFirst = 36 Hz and nLast = 2 ´ nFirst = 72 Hz
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