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A tuning fork gives 4beats with 50cm length of a Sonometer wire. If the length of the wire is shortened by 1cm, then the number of beats is still the same. Then, the frequency of the fork is:
(A) 396Hz
(B) 400Hz
(c) 404Hz
(D) 384Hz

Answer
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Hint: We shall use the basic property of a tuning fork, that is, the frequency of the tuning fork is inversely proportional to the length of the string. Since, the number of beats is the same before and after making a change in the length of the string, it means the change in frequency for beats to be observed is the same.

Complete answer:
Let the frequency of the tuning fork be fand let the frequency initially for 4 beats be f1 .
Then, it has been given to us in the question that:
ff1=4
f1=f4 [Let this expression be equation number (1)]
Let the frequency after cutting the wire by 1cm be f2 .
We know that the frequency of the tuning fork is inversely proportional to the length of the wire, that is:
f1l
fl=constant
Now, in the second case the length has been decreased, this implies that the frequency has increased.
As it's been given in the problem that, the number of beats is the same as 4. Therefore:
f2f=4
f2=f+4 [Let this expression be equation number (2)]
From equation number (1) and (2), we have:
f1l1=f2l2
Where, it has been given that:
 l1=50cml2=49cm
Putting the values of all these terms in the above equation, we get the frequency of tuning fork as:
(f4)50=(f+4)4950f49f=200+196f=396Hz
Hence, the frequency of the tuning fork comes out to be 396Hz.

Hence, option (A) is the correct option.

Note:
Our conclusion of the fact that the frequency afterwards was increased was based on a simple property that, length of wire is inversely proportional to frequency. These are some key properties regarding tuning forks. If we had not known this property and assumed the number of beats is the same so the frequency must be the same, then we might have ended up solving an equation which had no result.
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