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(D) Let the frequencies of the $10$ tuning forks be $f_1, f_2, \dots, f_{10}$ in increasing order.Since any two consecutive tuning forks produce $4$ b

Vedclass · Accepted Answer

$72$ and $36$

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(D) Let the frequencies of the $10$ tuning forks be $f_1, f_2, \dots, f_{10}$ in increasing order.Since any two consecutive tuning forks produce $4$ beats per second,the difference between consecutive frequencies is $4 \ Hz$.Thus,$f_n = f_1 + (n-1)d$,where $d = 4 \ Hz$.For $n = 10$,$f_{10} = f_1 + (10-1) 	imes 4 = f_1 + 36$.Given that the highest frequency is twice the lowest,$f_{10} = 2f_1$.Substituting the expression for $f_{10}$,we get $2f_1 = f_1 + 36$,which implies $f_1 = 36 \ Hz$.Then,$f_{10} = 2 	imes 36 = 72 \ Hz$.Therefore,the highest and lowest frequencies are $72 \ Hz$ and $36 \ Hz$,which corresponds to option $(d)$.

Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce $4$ beats per second. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies (in $Hz$) are ................

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