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# A tuning fork $A$ of frequency $512\,Hz$ produces $5$ beats per second when sounded with another tuning fork $B$ of unknown frequency. If $B$ is located with wax the number of beats is again $5$ per second. The frequency of the tuning fork $B$ before it was loaded is (A) $502\,Hz$(B) $507\,Hz$(C) $517\,Hz$(D) $522\,Hz$

Last updated date: 13th Sep 2024
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Answer
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HintThe frequency of $B$ is equal to the sum of the frequency of the $A$ and the number of beats produced from the frequency of $A$. And then the condition is given, the $B$ is located in the wax and again $5$ per second. By sing this condition, the frequency of $B$ is determined.

Complete step by step solution
Given that,
The tuning fork $A$ of frequency is, $n = 512\,Hz$,
The tuning fork of $A$ produces the beats of, $m = 5$.
Then the tuning fork $B$ is placed in the wax and again $5$ beats per second.
Now, the frequency of $B$ is given by,
$n' = n \pm m\,.....................\left( 1 \right)$
Where, $n'$ is the frequency of the tuning fork $B$.
By substituting the frequency of $A$ and the number of beats produced by the frequency of $A$ in the equation (1), then
$n' = 512 \pm 5$
By adding and subtracting the terms in the above equation, then the above equation is written as,
$n' = 517$ or $n' = 507$
On loading the tuning fork $B$, the frequency decreases from $517$ to $507$, so that the number of beats per second remains $5$.
Thus, the frequency of the tuning fork $B$ before it was loaded is $517\,Hz$.

Hence, the option (C) is the correct answer.

Note A tuning fork $A$ of frequency $512\,Hz$ produces $5$ beats per second when sounded with another tuning fork $B$ of unknown frequency. If $B$ is located with wax the number of beats is again $5$ per second. The frequency of the tuning fork $B$ before it was loaded is $517\,Hz$.