MHT CET · Physics · Waves and Sound
41 tuning forks are arranged in increasing order of frequency such that each produces 5 beats/second with next tuning fork. If frequency of last tuning fork is double that of frequency of first fork. Then frequency of first and last fork is
- A \(400,200 \mathrm{~Hz}\)
- B \(200,400 \mathrm{~Hz}\)
- C \(100,200 \mathrm{~Hz}\)
- D \(205,410 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(200,400 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Let Frequency of \(1^{\text {st }}\) tuning fork be \(=\mathrm{n}_1\)
\(\therefore \quad\) frequency of \(41^{\text {st }}\) tuning fork \(=n_{41}\) Now,
\(
\begin{array}{ll}
& \mathrm{n}_{41}=\mathrm{n}_1+(41-1) \times 5 \\
& \text { But, } \mathrm{n}_{41}=2 \mathrm{n}_1 \\
\therefore \quad & 2 \mathrm{n}_1=\mathrm{n}_1+200 \\
\therefore \quad & \mathrm{n}_1=200 \mathrm{~Hz} \\
\therefore \quad & \mathrm{n}_{41}=400 \mathrm{~Hz}
\end{array}
\)
\(\therefore \quad\) frequency of \(41^{\text {st }}\) tuning fork \(=n_{41}\) Now,
\(
\begin{array}{ll}
& \mathrm{n}_{41}=\mathrm{n}_1+(41-1) \times 5 \\
& \text { But, } \mathrm{n}_{41}=2 \mathrm{n}_1 \\
\therefore \quad & 2 \mathrm{n}_1=\mathrm{n}_1+200 \\
\therefore \quad & \mathrm{n}_1=200 \mathrm{~Hz} \\
\therefore \quad & \mathrm{n}_{41}=400 \mathrm{~Hz}
\end{array}
\)
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