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Standard X

Physics

Apparent Frequency

Tuning fork A...

Question

Tuning fork A of frequency $258 H z$ gives eight beats with a tuning fork $B$ . When prongs of $B$ are cut and again $A$ and $B$ are sounded, the number of beats heard remains the same. The frequency of $B$ is?

242 H z

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258 H z

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264 H z

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250 H z

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Solution

The correct option is A $242 H z$
$| F_{A} - F_{B} | = 8$
$| 258 - F_{B} | = 8$
$F_{B} = 250 / F_{B} = 266$
When B prongs cut $f π$ equation increases so that means initially $2 +$ was = $250$
So later become $266$
So initially $| 258 - 250 | = 8$
so later $| 258 - 266 | = 8$
so $f π$ eq of B is = $250$

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Tuning fork A of frequency 258 Hz gives eight beats with a tuning fork B. When prongs of B are cut and again A and B are sounded, the number of beats heard remains the same. The frequency of B is?

The correct option is A 242 Hz|FA−FB|=8|258−FB|=8FB=250/FB=266When B prongs cut fπ equation increases so that means initially 2+ was = 250So later become 266So initially |258−250|=8so later |258−266|=8so fπ eq of B is = 250

Tuning fork A of frequency $258 H z$ gives eight beats with a tuning fork $B$ . When prongs of $B$ are cut and again $A$ and $B$ are sounded, the number of beats heard remains the same. The frequency of $B$ is?