A) \[{{10}^{-5}}m\] and \[2.1\times {{10}^{13}}{{s}^{-1}}\]
B) \[1.4\times {{10}^{-6}}m\] and \[1.5\times {{10}^{12}}{{s}^{-1}}\]
C) \[{{10}^{-5}}m\]and \[1.5\times {{10}^{12}}{{s}^{-1}}\]
D) \[{{10}^{-5}}\]and \[1.32\times {{10}^{12}}{{s}^{-1}}\]
Correct Answer: A
Solution :
Idea This question is based on electromagnetic wave. Here, students must know about how to find frequency, speed and wavelength of a electromagnetic wave. Standard equation of electromagnetic wave is \[E={{E}_{0}}\sin ({{k}_{0}}x+{{\omega }_{0}}t)\] So, \[{{k}_{0}}=44\times {{10}^{4}}{{m}^{-1}}\] and \[{{\omega }_{0}}=132\times {{10}^{12}}{{s}^{-1}}\] Now \[{{k}_{0}}=\frac{2\pi }{\lambda }\Rightarrow 2\times \frac{22}{7}\times \frac{1}{{{\lambda }_{0}}}=44\times {{10}^{4}}\] \[\Rightarrow \] \[{{\lambda }_{0}}=\frac{1}{7}\times {{10}^{-4}}m\] and \[{{\omega }_{0}}=132\times {{10}^{12}}{{s}^{-1}}\] \[\Rightarrow \] \[2\pi {{v}_{0}}=132\times {{10}^{12}}\] \[\Rightarrow \]\[2\times \frac{22}{7}\times {{v}_{0}}=132\times {{10}^{12}}\] \[\Rightarrow \]\[{{v}_{0}}=21\times {{10}^{13}}{{s}^{-1}}\] Now in a medium of\[\mu =2\]only the\[{{\lambda }_{0}}\]will change to\[\lambda '=\frac{{{\lambda }_{0}}}{\mu }\]but no change to\[{{v}_{0}}\]shall occur. So, \[\lambda '=\frac{{{\lambda }_{0}}}{1.4}=\frac{{{10}^{-4}}}{1.4\times 7}=\frac{{{10}^{-4}}}{9.8}={{10}^{-5}}\]m and\[{{v}_{0}}=2.1\times {{10}^{13}}{{s}^{-1}}\] TEST Edge Questions on displacement current and about the nature of EM wave and about the permittivity and permeability of free space (or a medium) could be asked.
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