Question

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

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

Answer 1

The correct option is A $72$ and $36$

Let the frequency of first tuning fork is x then then we add on $4$ to x to get frequency of next fork so by doing this the frequency of the last fork will be $36+x.$

Now since it is given that the frequency of last tuning fork is double than that of first we get other frequency of last tuning fork =$2x$

Now by solving these two equations we get

$x=36$

therefore least frequency is $36$ and max. is $72$.