The frequency of two tuning forks $\mathrm{A}$ and $\mathrm{B}$ are $1 \cdot 5 \%$ more and $2 \cdot 5 \%$…

The frequency of two tuning forks $\mathrm{A}$ and $\mathrm{B}$ are $1 \cdot 5 \%$ more and $2 \cdot 5 \%$ less than that of the tuning fork $\mathrm{C}$. When $\mathrm{A}$ and $\mathrm{B}$ are sounded together, 12 beats are produced in 1 second. The frequency of tuning fork $\mathrm{C}$ is
  1. $200 \mathrm{~Hz}$
  2. $300 \mathrm{~Hz}$
  3. $240 \mathrm{~Hz}$
  4. $360 \mathrm{~Hz}$

Solution

$\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{c}} \times 1.015 \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{c}} \times 0.975$ $\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}=12$ $(1.015-0.975) \mathrm{f}_{\mathrm{c}}=12$ $0.040 \mathrm{f}_{\mathrm{c}}=12$ $\therefore \mathrm{f}=\frac{12}{0.040}=300 \mathrm{~Hz}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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