A) 10%
B) 20%
C) 21%
D) 5%
Correct Answer: C
Solution :
[c] \[K.E=\frac{{{L}^{2}}}{2M}\] \[{{L}_{1}}={{L}_{0}}{{L}_{2}}={{L}_{0}}+\frac{10{{L}_{0}}}{100}\] \[{{L}_{1}}={{L}_{0}}{{L}_{2}}+\frac{11{{L}_{0}}}{10}\] \[{{K}_{1}}=\frac{L_{0}^{2}}{2M}{{K}_{2}}=\frac{121L_{0}^{2}}{100\times 2M}\] \[\Delta K=\frac{L_{0}^{2}}{2M}\left[ \frac{121}{100}-1 \right]=\frac{21}{100}\frac{L_{0}^{2}}{2M}\] \[\frac{\Delta K}{K}%\frac{\left[ \frac{121}{100}-1 \right]\frac{L_{0}^{2}}{2M}}{\frac{L_{0}^{2}}{2M}}\times 100\] \[=21%\] Note: % change in quantity \[\left( \Delta K=\frac{{{K}_{2}}-{{K}_{1}}}{{{K}_{1}}} \right)\]
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