A) n
B) \[{{n}^{2}}\]
C) \[{{n}^{3}}\]
D) None of these
Correct Answer: B
Solution :
The equation of any normal to \[{{e}_{\max }}=0.005\times 10\times 100\pi =5\pi \] is \[{{v}_{0}}=\frac{1}{2\pi \sqrt{LC}}\] \[=\frac{1}{2\pi \sqrt{1\times {{10}^{-3}}\times 0.1\times {{10}^{-6}}}}\] \[=\frac{{{10}^{5}}}{2\pi }Hz\] ...(i) The straight line lx + my ? n = 0 will be normal to the hyperbola \[E=\frac{12375}{5000}\] then Eq. (i) and lx + my ? n = 0 represent the same line. \[=2.475eV=4\times {{10}^{-19}}J\] \[{{\text{l}}_{\text{eye}}}\text{=}\left( \text{photon flux} \right)\text{ }\!\!\times\!\!\text{ energy of a photon}\] \[{{\text{l}}_{\text{eye}}}\text{=}\left( 5\times {{10}^{4}} \right)\times 4\times {{10}^{-19}}\] \[=2\times {{10}^{-14}}\left( W/{{m}^{2}} \right)\] and \[=\frac{{{l}_{ear}}}{{{l}_{eye}}}=\frac{{{10}^{-13}}}{2\times {{10}^{-14}}}=5\] \[\mu =\frac{\Delta {{V}_{p}}}{\Delta {{V}_{g}}}\] \[\Rightarrow \] \[\Delta {{V}_{p}}=-\mu \times \Delta {{V}_{s}}\] \[=-50\left( -20 \right)=10V\] \[E=\Delta m{{c}^{2}}\] \[=0.5\times {{10}^{-3}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}\] \[=4.5\times {{10}^{13}}\] \[E=\frac{4.5\times {{10}^{13}}}{3.6\times {{10}^{6}}}\]
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