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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

80 and 40

B

100 and 50

C

44 and 22

D

36 and 72

Text Solution

Verified by Experts

The correct Answer is:
D
`N = (n_(L) - n_(f))/(x) + 1`
`10 -1 = (2n_(f) - n_(f))/(4)`
`9 xx 4 = n_(f)`
`n_(f) = 36 Hz`
`n_(1) = 2n_(f) = 2 xx 36 = 72 Hz`
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