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

Options

A. 80 and 40
B. 100 and 50
C. 44 and 22
D. 72 and 36

Answer

D

Explanation

Uning `n_(Last)=n_(first)+(N-1)x`
where `N=`number of tuning fork in series
`x=` beat frequency between two successive forks
`implies2n=n+(10-1)xx4impliesn=36Hz`
`:.n_("First")=36Hz` and `n_("Last")=2xxn_("First")=72Hz`