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Question

When two tuning forks, fork A and fork B are sounded together, 4 beats per second are heard. When the tuning fork B is loaded with wax then 6 beats per second are heard. If the frequency of fork A is 200 Hz then what was the original frequency of fork B?

A
209 Hz
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B
196 Hz
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C
202 Hz
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D
200 Hz
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Solution

The correct option is B 196 Hz
Given,
fb=4 Hz
We know, beat frequency
fb=|f1f2|
f2=f1±fb
Frequency of tuning fork A,
f1=200 Hz
f2=200±4 Hz
So, frequency of tuning fork B can be,
f2=204 Hz or 196 Hz
When tuning fork B is loaded with wax then,
fb=6 Hz
f2=200±6 Hz
So, frequency of tuning fork B can be,
f2=206 Hz or 194 Hz

As the tuning fork B is loaded with wax, therefore its time period should increase and frequency should decrease.
So, by intuition, the original frequency should be 196 Hz which will decrease to 194 Hz when loaded with wax.

OR

Given,
fb=4 Hz
We know, beat frequency
fb=|fAfB|=4 Hz
On waxing B, its frequency reduces and beat requency increases.
Hence,
fAfB=4 Hz
200fB=4 Hz
fB=196 Hz

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