Class 12th

Magnitude of electric field is constant & the surface is equipotential.

$\frac{\text{Energy}}{\text{Volume}}=\frac{{\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}}{{\mathrm{L}}^3}=\left({\mathrm{M}\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right)$

The distance between two successive bright fringes is fringe width  $\left(\beta \right)$ .

$\beta =\frac{\lambda D}{d}=\frac{589*{10}^{-9}*1.5}{0.15*{10}^{-3}}=5.9\mathrm{m}\mathrm{m}$

Relation between De-Broglie wavelength and K.E. is

$\lambda =\frac{\mathrm{h}}{\sqrt[]{2\left(\mathrm{K}\mathrm{E}\right){\mathrm{m}}_{\mathrm{e}}}}\Rightarrow \lambda \propto \frac{1}{\sqrt[]{\mathrm{K}\mathrm{E}}}$

$\begin{array}{rr}& \frac{{\lambda }_{\mathrm{A}}}{{\lambda }_{\mathrm{B}}}=\frac{\sqrt[]{{\mathrm{K}\mathrm{E}}_{\mathrm{B}}}}{\sqrt[]{\mathrm{K}\mathrm{E}}}\Rightarrow \frac{1}{2}=\sqrt[]{\frac{{\mathrm{T}}_{\mathrm{A}}-1.5}{{\mathrm{T}}_{\mathrm{A}}}}\\ & \Rightarrow {\mathrm{T}}_{\mathrm{A}}2\mathrm{e}\mathrm{V}\\ & \therefore {\mathrm{K}\mathrm{E}}_{\mathrm{B}}=2-1.5=0.5\mathrm{e}\mathrm{V}\\ & {?}_{\mathrm{B}}=4.5-0.5=4\mathrm{e}\mathrm{V}\end{array}$

$\mathrm{}\stackrel{?}{E}=\frac{\sigma }{2{\epsilon }_0}\mathrm{c}\mathrm{o}\mathrm{s}?{60}^{?}(-\hat{x})+\left. [\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}\mathrm{s}\mathrm{i}\mathrm{n}?{60}^{?}\right]\left(\hat{y}\right)$

$\stackrel{?}{\mathrm{E}}=\frac{\sigma }{2{\epsilon }_0}\left. [\left(1-\frac{\sqrt[]{3}}{2}\right)\stackrel{\mathrm{ˆ}}{\mathrm{y}}-\frac{1}{2}\stackrel{\mathrm{ˆ}}{\mathrm{x}}\right]$

$\mathrm{}I=I_0{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta$

$\frac{{\mathrm{I}}_0}{10}={\mathrm{I}}_0{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta$

$\mathrm{c}\mathrm{o}\mathrm{s}?\theta =\frac{1}{\sqrt[]{10}}=0.31<\frac{1}{\sqrt[]{2}}$ which is 0.707

So, $\theta >{45}^{?}$ ${}^{}$ and $90-\theta <{45}^{?}$ ${}^{}$ so only one option is correct i.e. ${18.4}^{?}$ ${}^{}$

Angle rotated should be $={90}^{?}-{71.6}^{?}={18.4}^{?}$ ${}^{}{}^{}{}^{}$ .

At any time $t$

$?=\mathrm{B}\mathrm{A}\mathrm{c}\mathrm{o}\mathrm{s}?\frac{\pi \mathrm{t}}{5}\mathrm{}\left(?\omega =\frac{2\pi }{\mathrm{T}}=\frac{2\pi }{10}=\frac{\pi }{5}\right)$

$\frac{\mathrm{d}?}{\mathrm{d}\mathrm{t}}=-\frac{\pi }{5}\mathrm{B}\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{\pi \mathrm{t}}{5}\right)$

$\mathrm{e}=\frac{\pi }{5}\mathrm{B}\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{\pi \mathrm{t}}{5}\right)$

e will be maximum when $\frac{\pi \mathrm{t}}{5}$ is $\frac{\pi }{2}$

$\mathrm{t}=2.5\mathrm{s}\mathrm{e}\mathrm{c}$

e will be minimum when $\frac{\pi \mathrm{t}}{5}$ is $\pi ,0$

$\mathrm{t}=\mathrm{0,5}\mathrm{s}\mathrm{e}\mathrm{c}$

So, answer will be (1).

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  $\frac{1}{f}=\left({\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{{\mathrm{f}}_{\mathcal{l}}}=\left(\frac{{\mu }_{\mathrm{g}}}{{\mu }_{\mathcal{l}}}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)$

(i) $\frac{{\mathrm{f}}_{\mathcal{l}}}{\mathrm{f}}=\frac{{\mu }_{\mathrm{g}}-1}{\frac{{\mu }_{\mathrm{g}}}{{\mu }_{\mathcal{l}}}-1}=\frac{1.5-1}{\frac{1.5}{1.41}-1}=8.875\approx 9$ .

First part of figure shown is or GATE and Second part of figure shown is NOT GATE So, $Y_P=OR+NOT=$  NOR GATE $Y=\stackrel{?}{A+B}=\bar{A}\cdot \bar{B}$

$\mathrm{}{\lambda }_{\text{electron}}=\frac{h}{\sqrt[]{2mE}}\mathrm{}\mathrm{}\frac{{\lambda }_{\text{electron}}}{{\lambda }_{\text{photon}}}=\frac{1}{c}\sqrt[]{\frac{E}{2m}}$

${\lambda }_{\text{photon}}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{E}}$