MHT CET · Physics · Waves and Sound
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
- A \(200 \mathrm{~Hz}\)
- B \(300 \mathrm{~Hz}\)
- C \(240 \mathrm{~Hz}\)
- D \(360 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(300 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{c}} \times 1.015 \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{c}} \times 0.975\)
\(\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}=12\)
\((1.015-0.975) \mathrm{f}_{\mathrm{c}}=12\)
\(0.040 \mathrm{f}_{\mathrm{c}}=12\)
\(\therefore \mathrm{f}=\frac{12}{0.040}=300 \mathrm{~Hz}\)
\(\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}=12\)
\((1.015-0.975) \mathrm{f}_{\mathrm{c}}=12\)
\(0.040 \mathrm{f}_{\mathrm{c}}=12\)
\(\therefore \mathrm{f}=\frac{12}{0.040}=300 \mathrm{~Hz}\)
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