Question

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency 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. 72 and 36

Answer 1

Answer: (D)

Using $n_{\text {Last }}=n_{\text {First }}+(N-1) x$
where $N=$ Number of tuning fork in series
$x=$ beat frequency between two successive forks
$\Rightarrow 2 n=n+(10-1) \times 4$
$\Rightarrow n=36\, Hz$
$\therefore n_{\text {First }}=36\, Hz$
and $n_{\text {Last }}=2 \times n_{\text {First }}=72\, Hz$