A) 4In2
B) \[\frac{\text{In}2}{2}\]
C) \[\frac{\text{In}2}{4}\]
D) 2In2
Correct Answer: C
Solution :
| \[R={{R}_{o}}^{e\,-\,\lambda t}\] |
| \[\therefore \] \[\frac{{{R}_{B}}}{{{R}_{A}}}=\frac{{{R}_{o}}e{{\,}^{-\,{{\lambda }_{{{B}^{t}}}}}}}{{{R}_{o}}{{e}^{-\,{{\lambda }_{{{B}^{t}}}}}}}\] |
| \[={{e}^{-({{\lambda }_{B}}-{{\lambda }_{A}})t}}={{e}^{-3t}}\] |
| \[\Rightarrow \] \[{{\lambda }_{B}}-{{\lambda }_{A}}=3\] \[\Rightarrow \] \[\frac{I{{n}^{2}}}{{{T}_{B}}}-\frac{In2}{In2}=3\]\[\Rightarrow \] \[{{T}_{B}}=\frac{In2}{4}\] |
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