MHT CET · Physics · Waves and Sound
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
- A 250 Hz
- B 300 Hz
- C 340 Hz
- D 400 Hz
Answer & Solution
Correct Answer
(A) 250 Hz
Step-by-step Solution
Detailed explanation
Let n be the frequency of fork C
\(\therefore \mathrm{n}_{\mathrm{A}}=\mathrm{n}+\frac{1.4 \mathrm{n}}{100}=\frac{101.4 \mathrm{n}}{100} \text { and } \mathrm{n}_{\mathrm{B}}=\mathrm{n}-\)\(\frac{2.6 \mathrm{n}}{100}=\frac{97.4 \mathrm{n}}{100}\)
But \(\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{B}}=10 \Rightarrow \frac{4 \mathrm{n}}{100}=10 \Rightarrow \mathrm{n}=250 \mathrm{~Hz}\)
\(\therefore \mathrm{n}_{\mathrm{A}}=\mathrm{n}+\frac{1.4 \mathrm{n}}{100}=\frac{101.4 \mathrm{n}}{100} \text { and } \mathrm{n}_{\mathrm{B}}=\mathrm{n}-\)\(\frac{2.6 \mathrm{n}}{100}=\frac{97.4 \mathrm{n}}{100}\)
But \(\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{B}}=10 \Rightarrow \frac{4 \mathrm{n}}{100}=10 \Rightarrow \mathrm{n}=250 \mathrm{~Hz}\)
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