A) \[{{T}^{2}}=k{{R}^{3}}\]
B) \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3/2}}={{\left( \frac{1.01R}{R} \right)}^{3/2}}\]
C) \[\Rightarrow \]
D) \[\frac{{{T}_{2}}}{{{T}_{1}}}={{(1+0.01)}^{3/2}}=1+\frac{3}{2}\times 0.01\]
Correct Answer: A
Solution :
Key Idea: Destructive interference is same as minimum intensity. The resultant intensity at a point due to two interfering waves \[\Delta {{i}_{C}}=0.96\times 7.2=6.91mA\] and \[\Delta {{i}_{E}}=\Delta {{i}_{C}}+\Delta {{i}_{B}}\] and having phase difference \[\therefore \] between them is given by \[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] For minimum intensity or destructive interference at a point \[=7.2-6.91=0.29mA\] that is \[\frac{(2n+1)\lambda }{2}\] \[\frac{(n+1)\lambda }{2}\] Also phase difference \[n(\lambda +1)\]path difference \[n\lambda \] path difference \[[M{{L}^{3}}{{T}^{-3}}]\] \[[M{{L}^{-1}}{{T}^{-1}}]\]
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