Question

The frequency of tuning forks $A$ and $B$ are respectively $3\%$ more and $2\%$ less than the frequency of tuning fork $C.$ When $A$ and $B$ are simultaneously excited, $5$ beats per second are produced. Then the frequency of the tuning fork $A($ (in $Hz)$ is

• A. $98$
• B. $100$
• C. $103$
• D. $105$

Answer 1

The correct option is C $103$

$$\begin{array}{c}Accordingtoquestion.....\\ n_A=n+\frac{3n}{100}=\frac{103n}{100}\\ n_B=n-\frac{2n}{100}=\frac{98n}{100}\\ Now,\\ n_A-n_{{}_B}=5\\ \Rightarrow \frac{103n}{100}-\frac{98n}{100}=5\\ \Rightarrow 5n=500\\ \Rightarrow n=100Hz\\ \therefore n_A=\frac{103n}{100}\\ =\frac{\left(103\right)\left(100\right)}{100}=103Hz\\ sothecorrectoptionisC.\end{array}$$