A) 44%
B) 40%
C) 20%
D) 10%
Correct Answer: A
Solution :
Speed of wave on a string \[v=\sqrt{\frac{T}{m}}\] or \[v\propto \sqrt{T}\] or \[\frac{{{v}_{2}}}{{{v}_{1}}}-\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\] or \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{v_{2}^{2}}{v_{1}^{2}}\] or \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\frac{v_{2}^{2}-v_{1}^{2}}{v_{1}^{2}}\] ... (i) Initially, \[{{T}_{1}}=120\,N,\,\,{{v}_{1}}=150\,m/s\] \[{{v}_{2}}={{v}_{1}}=\frac{20}{100}{{v}_{1}}={{v}_{2}}+\frac{{{v}_{1}}}{5}=\frac{6{{v}_{1}}}{5}\] \[=\frac{6}{5}\times 150=180\,m/s\] So, from Eq. (i), we get \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\frac{{{(180)}^{2}}-{{(150)}^{2}}}{{{(150)}^{2}}}\] \[=\frac{30\times 330}{150\times 150}=0.44\] Hence, % increase in tension \[=\left( \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}} \right)\times 100\] \[=0.44\times 100=44%\]
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