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The frequencies of three tuning forks A, B and C are related as $\mathrm{n}_{\mathrm{A}}>\mathrm{n}_{\mathrm{B}}>\mathrm{n}_{\mathrm{C}}$. Which the forks $\mathrm{A}$ and $\mathrm{B}$ are sounded together, the number of beats produced per second is ' $\mathrm{n}_1$ '. When forks $\mathrm{A}$ and $\mathrm{C}$ are sounded together the number of beats produced per second is ' $\mathrm{n}_2$ '. How may beats are produced per second when forks B and $\mathrm{C}$ are sounded together?
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Verified Answer
The correct answer is:
$\mathrm{n}_2-\mathrm{n}_1$
$$
\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{B}}=\mathrm{n}_1
$$
$$
\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{C}}=\mathrm{n}_2
$$
Subtracting Eq. (i) from eq. (ii)
$$
\mathrm{n}_{\mathrm{B}}-\mathrm{n}_{\mathrm{C}}=\mathrm{n}_2-\mathrm{n}_1
$$
\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{B}}=\mathrm{n}_1
$$
$$
\mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{C}}=\mathrm{n}_2
$$
Subtracting Eq. (i) from eq. (ii)
$$
\mathrm{n}_{\mathrm{B}}-\mathrm{n}_{\mathrm{C}}=\mathrm{n}_2-\mathrm{n}_1
$$
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