A) \[=v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\]
B) \[=v{{\left( \frac{3}{4} \right)}^{\frac{1}{2}}}\]
C) \[>v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\]
D) \[<v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\]
Correct Answer: C
Solution :
[c] \[hv_{0}^{2}-h{{v}_{0}}=\frac{1}{2}m{{v}^{2}}\] \[\therefore \frac{4}{3}h{{v}_{0}}-h{{v}_{0}}=\frac{1}{2}mv{{'}^{2}}\] \[\therefore \frac{v{{'}^{2}}}{{{v}^{2}}}=\frac{\frac{4}{3}v-{{v}_{0}}}{v-{{v}_{0}}}\text{ }\therefore v'=v\sqrt{\frac{\frac{4}{3}v-{{v}_{0}}}{v-{{v}_{0}}}\text{ }}\] \[\therefore \,\,\,v'>V\sqrt{\frac{4}{3}}\]
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