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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

`40 and 80`

B

`50 and 100`

C

`22 and 44`

D

`36 and 72`

Text Solution

Verified by Experts

The correct Answer is:
D
In ten forks , there are nine intervals
`n_(2) = n_(1) + 9 xx 4` (Also given `n_(2) = 2n_(1)`)
`2 n_(1) = n_(1) + 36`
`n_(1) = 36 Hz`
So `n_(2) = 2 n_(1) = 72 Hz`
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