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 \[{{T}_{2}}={{T}_{1}}+\frac{21}{100}{{T}_{1}}=1.21{{T}_{1}}\] ?. (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) in Eq. (iii), we get, \[1.1{{n}_{1}}-{{n}_{1}}=10\] \[0.1{{n}_{1}}=10\] \[{{n}_{1}}=100\,Hz\]
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