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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
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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
Solution:
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Verified Answer
The correct answer is:
$300 \mathrm{~Hz}$
$\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{c}} \times 1.015 \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{c}} \times 0.975$
$\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}=12$
$(1.015-0.975) \mathrm{f}_{\mathrm{c}}=12$
$0.040 \mathrm{f}_{\mathrm{c}}=12$
$\therefore \mathrm{f}=\frac{12}{0.040}=300 \mathrm{~Hz}$
$\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}=12$
$(1.015-0.975) \mathrm{f}_{\mathrm{c}}=12$
$0.040 \mathrm{f}_{\mathrm{c}}=12$
$\therefore \mathrm{f}=\frac{12}{0.040}=300 \mathrm{~Hz}$
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