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Ten tuning forks are arranged in increas...

Ten tuning forks are arranged in increasing order of frequency is such a way that any two nearest tuning forks produce `4 beast//sec`. The highest freqeuncy is twice of the lowest. Possible highest and the lowest frequencies are

A

72, 144

B

36, 72

C

18, 36

D

9, 18

Text Solution

Verified by Experts

The correct Answer is:
B
`N = (n_(L) - n_(F))/(x) +1`
`10 = (2n_(F) - n_(F))/(4) + 1`
`9 xx 4 = n_(F)`
`n_(F) = 36`
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