A) \[44%\]
B) \[40%\]
C) \[20%\]
D) \[10%\]
Correct Answer: A
Solution :
We know that\[v=\sqrt{\frac{T}{m}}\] \[v\propto \sqrt{T}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}\,=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}={{\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)}^{2}}\] \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\frac{v_{1}^{2}-v_{1}^{2}}{v_{1}^{2}}\] \[{{v}_{2}}={{v}_{1}}+\frac{20}{100}{{v}_{1}}=\frac{120}{100}{{v}_{1}}=\frac{6}{5}{{v}_{1}}\] \[=\frac{6}{5}\times 150\] \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\frac{{{(180)}^{2}}-{{(150)}^{2}}}{{{(150)}^{2}}}\] \[=\frac{30\times 330}{150\times 180}=0.47\] \[=\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}\times 100=0.44\times 100=44%\]
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