Class 12th

$\begin{array}{l}{\displaystyle {\int }_2^3\frac{1}{x}dx}={\displaystyle {\int }_{-2}^3\frac{1}{x}dx}={\left[logx\right]}_2^3\\ =log3–log2\\ =log\frac{3}{2}\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) When the A stops moving the current in B become zero, it is possible only if the current in A is constant. If the current in A would be variable, there must be an induced emf (current) in B even if the A stops moving.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, c, d)

Explanation- Electrostatic shielding is the phenomenon to block the effects of an electric field. The conducting shell can block the effects of an external field on its internal content or the effect of an internal field on the outside environment.

Magnetostatic shielding is done by using an enclosure made of a high permeability

magnetic material to prevent a static magnetic field outside the enclosure from reaching

objects inside it or to confine a magnetic field with in the enclosure.

This is a multiple choice answer as classified in NCERT Exemplar

(b) When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, d)

Explanation- for solenoid H=nI

H= 1000 $*$ 1= 1000 Am

Thus H is a constant, so it is nearly unchanged

B= ${\mu }_0{\mu }_r$ n I

So B $\propto$ ${\mu }_r$

( $\aleph$ )_m)_fero = 10³

( $\aleph$ )_m)_para = 10^-5

$\frac{\left(\right(\aleph \left)\mathrm{m}\right)\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{o}}{\left(\right(\aleph \left)\mathrm{m}\right)\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}}$ = $\frac{{10}^3}{{10}^{-5}}$ = 10⁸

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, d)

Explanation- The primary origin of magnetism lies in the fact that the electrons are revolving and spinning of an atom, which gives rise to current called atomic current.

This atomic current gives rise to magnetism. The revolving and spinning about nucleus of

An atom is called intrinsic spin of electron.

This is a multiple choice answer as classified in NCERT Exemplar

$\varnothing =B.A$

And A= A₁+A₂= (L²k+L²i)

B= B_o (i+k)T

$\varnothing =B.A$= B_o (i+k) (L²k+L²i)

= 2B_oL²Wb

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation – magnetic field lines for magnetic induction B form continuous lines. So lines of B are necessarily continuous across S.

Also H varies inside and outside. So lines of H cannot all be continuous across S

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- as curie law explains B $\propto \frac{\mathit{B}\left(\mathit{m}\mathit{a}\mathit{g}\mathit{n}\mathit{e}\mathit{t}\mathit{i}\mathit{c}\mathit{}\mathit{f}\mathit{i}\mathit{e}\mathit{l}\mathit{d}\mathit{}\mathit{i}\mathit{n}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\right)}{\mathit{t}\left(\mathit{t}\mathit{e}\mathit{m}\mathit{p}\mathit{}\mathit{i}\mathit{n}\mathit{}\mathit{k}\mathit{e}\mathit{l}\mathit{v}\mathit{i}\mathit{n}\right)}$

$\frac{I_2}{I_1}=\frac{B_2}{B_1}*\frac{t_1}{t_2}$

_{$I$ 1} =8Am^-1, B₁= 0.6T, t₁=4K

B₂= 0.2T, t₂=16K

So after solving we get I₂= 2/3Am^-

$\begin{array}{l}{\displaystyle {\int }_{-1}^1\left(x+1\right)dx}={\displaystyle {\int }_{-1}^1.xdx+{\displaystyle {\int }_{-1}^{1}dx}}\\ ={\left[\frac{x^2}{2}\right]}_{-1}^1+{\left[x\right]}_{-1}^1\\ =\left[\frac{1^2}{2}-\frac{{\left(-1\right)}^2}{2}\right]+\left[1-\left(-1\right)\right].\\ =\left[\frac{1}{2}-\frac{1}{2}\right]+\left[1+1\right]=2.\end{array}$