Class 12th

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

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

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

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

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

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 $v_y=v_0$

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

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

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

Kindly consider the following Image

Kindly consider the following Image