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
$\Rightarrow 2n=n+(10-1)\times 4$
$\Rightarrow n=36Hz$
$\therefore n_{\text{First }}=36Hz$
and $n_{\text{Last }}=2\times n_{\text{First }}=72Hz$