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Standard XII

Physics

Ropes

A Tuning fork...

Question

A Tuning fork ‘A’ produces 6 beats per second with another fork ‘B’. On loading ‘B’ with a little wax, it produces 5 beats per second with ‘A’. If the frequency of ‘A’ is 256 Hz, the frequency of ‘B’ is:

262 Hz

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250 Hz

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256 Hz

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261 Hz

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Solution

The correct option is A 262 Hz
On loading wax on $B$ frequency of $B$ decreases and beat also decreases.
So, frequency of $B$ is greater than frequency of $A$ .
$B e a t s = δ_{B} - δ_{A}$
$6 = δ_{B} - 256$
$\Rightarrow δ_{B} = 262 H_{z}$

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Similar questions

Q. A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?

A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?

Q. A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and number of beats heard are 6 per second. The original frequency of the tuning fork is

Q. A tuning fork produces

4

beats per second, with another tuning fork of frequency

256 Hz

. The first one is now loaded with a little wax and the beat frequency is found to increase to

6

per second. The original frequency

(in Hz)

of the tuning fork was _____.

Q. A tuning fork

A

of unknown frequency produces

4

beats per second when sounded with another tuning fork

B

of frequency

254 Hz

. When tuning fork

A

is loaded with wax, it still gives

4

beats per second with tuning fork

A

. The initial unknown frequency of tuning fork

A

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A Tuning fork ‘A’ produces 6 beats per second with another fork ‘B’. On loading ‘B’ with a little wax, it produces 5 beats per second with ‘A’. If the frequency of ‘A’ is 256 Hz, the frequency of ‘B’ is:

The correct option is A 262 HzOn loading wax on B frequency of B decreases and beat also decreases. So, frequency of B is greater than frequency of A.Beats=δB−δA6=δB−256 ⇒δB=262 Hz

The correct option is A 262 Hz
On loading wax on $B$ frequency of $B$ decreases and beat also decreases.
So, frequency of $B$ is greater than frequency of $A$ .
$B e a t s = δ_{B} - δ_{A}$
$6 = δ_{B} - 256$
$\Rightarrow δ_{B} = 262 H_{z}$