A) 0.01
B) 0.02
C) 0.03
D) 0.04
Correct Answer: B
Solution :
Beats per second when both the wires vibrate simultaneously. \[{{n}_{1}}\pm {{n}_{2}}=6\] or \[\frac{1}{2l}\sqrt{\frac{T}{m}}\pm \frac{1}{2l}\sqrt{\frac{T}{m}}=6\] or \[\frac{1}{2l}\sqrt{\frac{T}{m}}-\frac{1}{2l}\sqrt{\frac{T}{m}}=6\] or \[\frac{1}{2l}\sqrt{\frac{T}{m}}-600=6\] \[\frac{1}{2l}\sqrt{\frac{T}{m}}-606\] ??(i) Given that fundamental frequency \[\frac{1}{2l}\sqrt{\frac{T}{m}}=600\] ?..(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{\frac{1}{2l}\sqrt{\frac{T}{m}}}{\frac{1}{2l}\sqrt{\frac{T}{m}}}=\frac{606}{600}\] or \[\sqrt{\frac{T}{T}}=(1.01)\] or \[\frac{T}{T}=(1.02)%\] or \[T=T(1.02)\] Increase in tension \[\Delta T=T\times 1.02-T\] \[=(0.02T)\] Hence, \[\Delta T=0.02\]You need to login to perform this action.
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