Class 12th

This is a long answer type question as classified in NCERT Exemplar

(i) $\left(i\right)\int E.dl$ = ${\int }_1^{2}E.dl+{\int }_2^{3}E.dl+{\int }_3^{4}E.dl+{\int }_4^{1}E.dl$

= E_oh[sin(kz₂-wt)-sin(kz₁-gwt)]

(ii) $\int B.ds=\int B.dscos0={\int }_{z1}^{z2}{B}_0\mathrm{s}\mathrm{i}\mathrm{n}?(kz-wt)hdz$

= $\frac{-B_{o}h}{k}\cos ?\left(kz2-wt\right)\cos ?\left(kz1-wt\right)$

(iv) $?E.dl=\frac{-d\varnothing }{dt}$ =- $?B.ds$

=Eoh[sin(kz2-wt)-sin(kz1-wt)]

=- $\frac{d}{dt}[\frac{B_{yh}}{k}\cos ?\left(kz2-wt\right)-\mathrm{c}\mathrm{o}\mathrm{s}?(kz1-wt\left)\right]$

Eo=Bow/k=Byc

Eo/Bo=c

This is a long answer type question as classified in NCERT Exemplar

E(s,t)= ${\mu }_o{I}_o\cos ?\left(2\pi vt\right)In\left(\frac{s}{a}\right)k$

Now displacement current J_d=e_{o $\frac{dE}{dt}=\frac{{\epsilon }_{o}d}{dt}\left[{\mu }_o{I}_o\cos ?\left(2\pi vt\right)In\left(\frac{s}{a}\right)k\right]$}

= $\frac{{\mu }_o{\epsilon }_o{I}_{o}vd}{dt}\left[cos2v\pi tIn\left(\frac{s}{a}\right)k\right]$

= $\frac{v^2}{c^2}2\pi I_{o}sin2\pi vtIn\left(\frac{a}{s}\right)k$

${\frac{2\pi a}{2\lambda })}^2{I}_0=({\frac{a\pi }{\lambda })}^2{I}_0$ = $\frac{2\pi I_0}{{\lambda }_2}$ ina/s sin2 $\pi vtk$

I_d= $\int J_{d}sdsd\theta ={\int }_0^1{J}_{s}sds{\int }_0^{2\pi }d\theta$

  =( $\frac{2\pi }{\lambda }$ )^{2 $I_o{\int }_0^a\frac{a}{s}sdssin\pi vt$}

= $\frac{a^2}{2}{\left(\frac{2\pi }{\lambda }\right)}^2{I}_{o}sin2\pi vt{\int }_0^{a}In\left(\frac{a}{s}\right).d({\frac{s}{a})}^2$

After solving this we get

I_d= $\frac{a^2}{4}({\frac{2\pi }{\lambda })}^2{I}_{0}sin2\pi vt$

I_d= ${\left(\frac{2\pi a}{2\lambda }\right)}^2{I}_{o}sin2\pi vt=I_{od}sin2\pi vt$

I_od= ( ${\frac{2\pi a}{2\lambda })}^2{I}_0=({\frac{a\pi }{\lambda })}^2{I}_0$

$\frac{I_{od}}{I_o}=({\frac{a\pi }{\lambda })}^2$

This is a long answer type question as classified in NCERT Exemplar

Suppose the distance between the plates is d and applied voltage is V_t= V₀2

Then electric field is E= $\frac{V}{d}$ sin(2 $\pi vt$ )

J_c= $\frac{1}{\rho }\frac{V_0}{d}sin\left(2\pi vt\right)$

  =  $J_0^{c}sin2\pi vt$

$J_0^c$   = $\frac{V_0}{\rho d}$

J_d= $\epsilon \frac{dE}{dt}$ = $\frac{V_0}{\rho d}$

   = $\epsilon \frac{2\pi v{V}_0}{d}$ cos(2 $\pi vt$ )

   = J₀^d cos 2 $\pi vt$

J₀^d= $\frac{2\pi V_{\epsilon }{V}_0}{d}$

$\frac{J_0^d}{J_0^c}$ = 2 $\pi V_{\epsilon }\rho$ = 2 $\pi 80{\epsilon }_0$ v $*$ 0.25 = 4 $\pi {\epsilon }_0$ v $*10$ =4/9

This is a long answer type question as classified in NCERT Exemplar

E= $\frac{\lambda e_s}{2\pi {\epsilon }_{0}a}j$

And B= $\frac{{\mu }_{0i}}{2\pi a}$ i= $\frac{{\mu }_{0\lambda v}}{2\pi a}$ i

Then S= $\frac{1}{{\mu }_0}$ {E $*B$ }= $\frac{1}{{\mu }_0}\{\frac{{\lambda }_j}{2\pi {\epsilon }_{0}a}*\frac{{\mu }_{0i}}{2\pi a}\mathrm{\lambda }\sqrt[]{i}\}$

 = $\frac{{\lambda }^{2}v}{4{\pi }^2{\epsilon }_{0a^2}}\left(j*i\right)=-\frac{{\lambda }^{2}v}{4{\pi }^2{\epsilon }_{0a^2}}k$

Yes, passing Class 12 is mandatory in order to secure admission to Techno India University. Students seeking for admission to UG or PG courses at the college must qualify for Class 12 with minimum 50-55% aggregate in Class 12.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-as we know that $\lambda =\frac{h}{\sqrt[]{2mk}}$

so $\lambda$ inversely proportional to $\sqrt[]{m}$

so from the above conclusion we can say that $\lambda$ _alpha $<\lambda$ _p $=\lambda$ _n $<\lambda$ _e

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (d)

explanation- When a beam of electrons of energy E₀ is incident on a metal surface kept in an

evacuated chamber electrons can be emitted with maximum energy E₀ (due to elastic

collision) and with any energy less than E₀, when part of incident energy of electron is

used in liberating the electrons from the surface of metal.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b)

Explanation- E=hc/ $\lambda$

And l = hc/E= 1240/10⁶= 1.24 $*$ 10^-3nm