NEET-XII-Physics
neet 2020 questions with solutions year:2020
- #12In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency `` 6\,Hz `` . When tension in B is slightly decreased, the be at frequency increases to `` 7\,Hz `` . If the frequency of A is `` 530\,Hz `` , the original frequency of B will be :
(A)523 Hz
(B)524 Hz
(C)536 H z
(D)537 H z
digAnsr: BAns : It is given, the difference of `` f_{A} `` and `` f_{B} `` is `` 6\,HZ ``
Guitar string i.e. string is fixed from both ends
Frequency `` \propto \sqrt{\text{Tension}} ``
If tension in B slightly decrease then frequency of B decreases
If B is 536 Hz, as the frequency decreases, beats with A also decreases
If B is 524 Hz, as the frequency decreases, beats with A increases
If tension decreases, `` f_{B} `` decreases and becomes `` f'_{B} ``
Now, difference of `` f_{A} `` and `` f'_{B}=7\,Hz `` (increases)
So, `` f_{A}=f_{B} ``
`` f_{A}-f_{B}=6\,Hz ``
`` f_{A}=530\,Hz ``
`` f_{B}=524\,Hz `` (original)