41 tuning forks are arranged in increasing order of frequency such that each produces 5 beats/second with…
41 tuning forks are arranged in increasing order of frequency such that each produces 5 beats/second with next tuning fork. If frequency of last tuning fork is double that of frequency of first fork. Then frequency of first and last fork is
$400,200 \mathrm{~Hz}$
$200,400 \mathrm{~Hz}$
$100,200 \mathrm{~Hz}$
$205,410 \mathrm{~Hz}$
Solution
Let Frequency of $1^{\text {st }}$ tuning fork be $=\mathrm{n}_1$
$\therefore \quad$ frequency of $41^{\text {st }}$ tuning fork $=n_{41}$ Now,
$\begin{array}{ll}
& \mathrm{n}_{41}=\mathrm{n}_1+(41-1) \times 5 \\
& \text { But, } \mathrm{n}_{41}=2 \mathrm{n}_1 \\
\therefore \quad & 2 \mathrm{n}_1=\mathrm{n}_1+200 \\
\therefore \quad & \mathrm{n}_1=200 \mathrm{~Hz} \\
\therefore \quad & \mathrm{n}_{41}=400 \mathrm{~Hz}
\end{array}$