A) \[\frac{1}{8}\]
B) \[\frac{1}{32}\]
C) less than \[\infty \]
D) greater than\[f(x)=\left\{ \begin{align} & \frac{{{\sin }^{3}}(\sqrt{3}).\log (1+3x)}{{{({{\tan }^{-1}}\sqrt{x})}^{2}}({{e}^{5\sqrt{x}}}-1)x},x\ne 0 \\ & a,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0 \\ \end{align} \right.\]
Correct Answer: D
Solution :
From\[5.00\pm 0.10cm\] (where \[{{m}_{A}}\]) If wavelength of incident light charges from \[{{r}_{A}}\]to \[{{r}_{B}}\](decreases). Let. energy of incident light charges from E to E' and speed of fastest electron changes from v to Ö then \[{{T}_{A}}\] ?(i) andÖ\[{{T}_{B}}\] ?(ii) As\[{{m}_{A}}\] Hence, Ö\[{{m}_{B}}\] \[\frac{{{T}_{A}}}{{{T}_{B}}}={{\left( \frac{{{r}_{A}}}{{{r}_{B}}} \right)}^{3/2}}\]Ö\[{{T}_{A}}>{{T}_{B}}(if\,{{r}_{A}}>{{r}_{B}})\] \[{{T}_{A}}>{{T}_{B}}(if\,{{m}_{A}}>{{m}_{B}})\]Ö\[{{T}_{A}}={{T}_{B}}\] So Ö\[\frac{2R}{\sqrt{15}}\]You need to login to perform this action.
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