Question

25 tunning forks are arranged in series in the order of decreasing frequency . Any two successive forks produce 3 beats/s. If the frequency of the first turning fork is the octave of the last fork, then the frequency of the ${21}^{st}$ fork is

• A. $72HZ$
• B. $288HZ$
• C. $84HZ$
• D. $87HZ$

Answer 1

The correct option is C $84HZ$

The tuning forks are arranged in decreasing frequencies. Therefore, as per given in the question frequencies of first and last tuning forks are 2nandn respectively.

Frequencies in the given arrangement are as follows

n1→2n
n2→2n−1×3
n3→2n−2×3
n4→2n−3×3
.
.
n25→2n−24×3

Imposing given condition we get

2n−24×3=n
⇒n=72

So, frequency of 21st tuning fork

n₂₁=2n−15×3
⇒n₂₁=2×72−20×3=84 Hz