A) \[t=\sqrt{2}s\]changes to \[a=\frac{dv}{dt}=\frac{d}{dt}(4{{t}^{3}}-2t)\]and \[a=12{{t}^{2}}-2\]changes to NO
B) \[t=\sqrt{2}\sec \]changes to \[a=12\times {{(\sqrt{2})}^{2}}-2=22m/{{s}^{2}}\] and NO changes to\[\therefore \frac{da}{dt}=k\]
C) \[\Rightarrow a=kt\] changes to \[\Rightarrow \] and NO changes to \[\frac{dv}{dt}=kt\Rightarrow dv=ktdt\]
D) No charge transfer takes place
Correct Answer: C
Solution :
\[\frac{20!}{{{(5!)}^{4}}}\] \[\frac{20!}{{{(4!)}^{5}}}\] Based on magnetic properties of the complex, it is found that iron has three unpaired electrons thus it exists as \[\frac{1}{14}\] formed by reduction of \[\frac{3}{14}\]by NO which is oxidised to \[{{\left( \sqrt{x}-\frac{k}{{{x}^{2}}} \right)}^{10}}\].You need to login to perform this action.
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