The frequency of two tuning forks A and B are respectively $1.4 \%$ more and $2.6 \%$ less than that of the…
The frequency of two tuning forks A and B are respectively $1.4 \%$ more and $2.6 \%$ less than that of the tuning fork C . When A and B are sounded together, 10 beats are produced in 1 second. The frequency of tuning fork C is
250 Hz
300 Hz
340 Hz
400 Hz
Solution
Let n be the frequency of fork C
$\therefore \quad \mathrm{n}_{\mathrm{A}}=\mathrm{n}+\frac{1.4 \mathrm{n}}{100}=\frac{101.4 \mathrm{n}}{100} \text { and } \mathrm{n}_{\mathrm{B}}=\mathrm{n}-\frac{2.6 \mathrm{n}}{100}=\frac{97.4 \mathrm{n}}{100}$ But $\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{B}}=10 \Rightarrow \frac{4 \mathrm{n}}{100}=10 \Rightarrow \mathrm{n}=250 \mathrm{~Hz}$