A) 44%
B) 40%
C) 20%
D) 10%
Correct Answer: A
Solution :
Speed of wave on a string \[v=\sqrt{\frac{T}{m}}\] where T is tension in the string and m is mass per unit length of string. Or \[v\propto \sqrt{T}\] Or \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] Or \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)}^{2}}\] Or \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\frac{v_{2}^{2}-v_{1}^{2}}{v_{1}^{2}}\] Given, \[{{T}_{1}}=120m,{{v}_{1}}=150m/s\] \[{{v}_{2}}={{v}_{1}}+\frac{20}{100}{{v}_{1}}=\frac{120}{100}{{v}_{1}}=\frac{6}{5}{{v}_{1}}=\frac{6}{5}\times 150\] \[=180\text{ }m/s\] Substituting the values in Eq. (i), we get \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\frac{{{(180)}^{2}}-{{(150)}^{2}}}{{{(150)}^{2}}}\] \[=\frac{30\times 330}{150\times 150}=0.44\] Per cent increase in tension, \[=\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}\times 100\] \[=0.44\times 100\] \[=44%\]
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