Physics Ncert Solutions Class 12th

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So,  ${}_{}\frac{}{}$ $V_{ab}=\frac{30}{5+10}*5=10\mathrm{V}$

$\mathrm{}50.00$

$x=$ retentivity

$y=$ coercivity

$\mathrm{z}=$ saturation magnetization

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).

Kindly consider the following answer

$\overrightarrow{\mu }=-\frac{e}{2m}\overrightarrow{L}$

 ${}_{}\frac{}{}$ $K.E_1=\frac{1230}{800}-?$ - (i)

${}_{}{}_{}\frac{}{}$  $K.E_2=2K.E_1=\frac{1230}{500}-?$ - (ii)

from (i) & (ii)

$$ $?=0.615$  ev

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

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

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

  $\frac{{\mu }_A}{{\mu }_B}=\frac{C/v_A}{C/v_B}=\frac{v_B}{v_A}=\frac{1}{2}$

Let the thickness is d

$d=\frac{5*10^{-10}*v_A{v}_B}{v_A-v_B}$          

d = 5 * 10^-10 * v_A

In primary rainbow, red colour is at top and violet is at bottom.

Intensity of secondary rainbow is less in comparison to primary rainbow.

Assuming current 'i' In outer loop magnetic field at centre

$=4*\frac{{\mu }_0}{4\pi }\frac{i}{L/2}*[2sin45°)$

$=\frac{2\sqrt[]{2}{\mu }_{0}i}{\pi L}$

M = $\frac{Fm\alpha throughinnerloop}{i}$ $\frac{}{}$

= $\frac{2\sqrt[]{2}{\mu }_{0}i{\mathcal{l}}^2}{\pi L}=\frac{2\sqrt[]{2}{\mu }_0{\mathcal{l}}^2}{\pi L}$ $\frac{\text{exception 25:}{}_{}{}^{}}{}\frac{\text{exception 25:}{}_{}{}^{}}{}$

Q = CV

$V=\frac{Q}{C}$

Straight line with slope = $\frac{1}{C\text{?}}$ $\frac{}{}$

$Slope=\frac{1}{C}=\frac{1}{2*10^{-6}}=5*10^5$