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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

144

B

36, 72

C

18, 36

D

9, 18

Text Solution

Verified by Experts

The correct Answer is:
b
Suppose f be the lowest frequency as beat frequency
is 4 between any two conseutive forces. The last fork will
have a frequency `= f + (9 xx 4 ) = f + 36`
Since, frequency of the last tuning fork is twice that of the
first
`2f= f+ 36 rArr f = 36`
Here, the frequency of last fork is
`2f= 2xx 36 = 72`
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