Class 12th

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

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

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

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

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

Yes, R3 Jewel Academy may accept Class 12th marks. Candidates seeking admission to all programmes are selected based on merit. Candidates who have also completed Class 12 can also apply for various courses. The college offers seven certificate courses, including Jewellery Designing, Advanced Jewellery CAD Matrix, Jewellery Key Shot Training, etc. R3 Jewel Academy admissions are based on merit.

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

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

(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 $Y=\stackrel{?}{A+B}=\bar{A}\cdot \bar{B}$

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

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

$\mathrm{}V=\frac{1}{\sqrt[]{\mu \epsilon }}n=\sqrt[]{{\mu }_r{\epsilon }_r}=2$

$\mathrm{s}\mathrm{i}\mathrm{n}?c=\frac{1}{2}c={30}^{?}$

Total power is $(15*45)+(15*100)+(15*10)+(2*1000)=4325\mathrm{W}$ .

So, current is $\frac{4325}{220}=19.66\mathrm{A}$

Answer is $20\mathrm{A}\mathrm{m}\mathrm{p}$ .

Since magnetic force cannot change the speed. So only electric field which is along $x$ -direction will change the speed along $X$ -direction only.

$v_x=\frac{E_{0}q}{m}t$ but $v_y=v_0$

$2v_0=\sqrt[]{v_x^2+v_y^2}$

$4v_0^2=\frac{E_o^2{q}^2{t}^2}{m^2}+v_0^2$

$t=\frac{\sqrt[]{3}m{v}_o}{q{E}_o}$