Physics

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

Microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules.

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

The orientation of the portable radio with respect to broadcasting station is important because the electromagnetic waves are plane polarised, so the receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave.

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

(i) U_E= $\frac{1}{2}{\epsilon }_o{E}^2$

U_B= $\frac{1}{2}\frac{B^2}{{\mu }_o}$

So total energy density = U_E+ U_B= $\frac{1}{2}{\epsilon }_o{E}^2$ + $\frac{1}{2}\frac{B^2}{{\mu }_o}$

E= Eo/ $\sqrt[]{2}$  and B=Bo/ $\sqrt[]{2}$

U_av= $\frac{1}{4}{\epsilon }_o{E}^2$ + $\frac{1}{4}\frac{B^2}{{\mu }_o}$

(ii) We know Eo=cBo and c = 1/ $\sqrt[]{{\mu }_0{\epsilon }_0}$

$\frac{1}{4}\frac{B^2}{{\mu }_o}=\frac{1}{4}\frac{\frac{{Eo}^2}{c^2}}{{\mu }_o}$ = 1/4 $\epsilon oEo$ ²

U_E= U_B

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

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

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