A) 210
B) 200
C) 110
D) 100
Correct Answer: D
Solution :
For transverse vibration in the string frequency is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\](where T is tension, m is the mass per unit length ) Hence, \[{{n}_{1}}=\frac{1}{2l}\sqrt{\frac{{{T}_{1}}}{m}}\] ?(i) and \[{{n}_{2}}=\frac{1}{2l}\sqrt{\frac{{{T}_{2}}}{m}}\] ?(ii) Number of beat \[{{n}_{2}}-{{n}_{1}}=10\] ?(iii) (given) Given: \[{{T}_{2}}={{T}_{1}}+\frac{21}{100}{{T}_{1}}=1.21{{T}_{1}}\] ?(iv) Putting the value of \[{{T}_{2}}\]in Eq.(ii), we get \[{{n}_{2}}=\frac{1}{2l}\sqrt{1.21\frac{{{T}_{1}}}{m}}=1.1\times \frac{1}{2l}\sqrt{\frac{{{T}_{1}}}{m}}\] \[{{n}_{2}}=1.1{{n}_{1}}\] ?(v) Putting the value of \[{{n}_{2}}\]from Eq. (v) Eq. (ii),we get \[1.1{{n}_{1}}-{{n}_{1}}=10\] \[0.1{{n}_{1}}=10\] \[{{n}_{1}}=100Hz\]You need to login to perform this action.
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