physics ncert exemplar solution class 12th chapter one

$E=\frac{kq}{2a^2}$

$E\text{'}=\frac{kq}{{\left(\sqrt[]{2}a\right)}^2}=\frac{kq}{2a^2}$

$E_{Net}=2Ecos45°+E\text{'}$

$=\frac{kq}{a^2}\left(\frac{1}{\sqrt[]{2}}+\frac{1}{2}\right)$

 ${\rho }_r=\{\begin{array}{cc}\hfill {\rho }_0\left(\frac{3}{4}-\frac{r}{R}\right)\hfill & \hfill forr\le R\hfill \\ \hfill 0\hfill & \hfill forr>R\hfill \end{array}$

$P_r=E4\pi r^2=\frac{q_{enc}}{{\epsilon }_0}$

As,  $E4\pi r^2=\frac{\pi {\rho }_0{r}^3}{{\epsilon }_0}\left(1-\frac{r}{R}\right)$

$E=\frac{{\rho }_{0}r}{4{\epsilon }_0}\left(1-\frac{r}{R}\right)$

Sol. $E=\left(I\right)\left(t\right)\left(A\right)?{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta ?$
$\Rightarrow \mathrm{}\left(3.3\right)\left. [\frac{2\pi }{31.4}\right]\left. [3*{10}^{-4}\right]*\frac{1}{2}$

$\Rightarrow \mathrm{}0.99*{10}^{-4}$

$F_{23}=\frac{k{q}_2{q}_3}{d^2}=\frac{k*2*2}{12}=4k$  

$F_{21}=4k$                                        

$F_{13}=4k$             

Net force on q₂ = F_R = $\sqrt[]{F_{21}^2+F_{13}^2+2F_{21}{F}_{23}cos\theta }$  

= $\sqrt[]{{\left(4k\right)}^2+{\left(4k\right)}^2+2*4kcos60°}$  

=  $k\sqrt[]{48}-\left(1\right)$

Force b/w (1) & (2) F₂₁ = 4k – (2)p

$\frac{F_R}{F_{21}}=\frac{k\sqrt[]{48}}{4k}=\frac{4\sqrt[]{3}}{4}=\sqrt[]{3}:1$

          

${?}_{flot}=\frac{q}{2{\in }_0}\left(1-cos\theta \right)$

When, 'q' is at the centre of the flat surface then, $\theta =50.$

${?}_{flat}=\frac{q}{2{\in }_0}\left(1-cos90°\right)$

$=\frac{q}{2{\in }_0}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{\rho l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{\rho l}{A}$ l²= 1.7 $*$ 10^{-8 $*\frac{10}{\pi *{10}^{-6}}*81$} j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000}*$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000}*$ 100= 0.32%

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- the positive charge Q is uniformly distributed at the outer surface of the enclosed sphere thus electric field inside the sphere is zero. So the effect of electric field on charge q due to positive charge Q is zero.

Now the only attractive and repulsive force between Q and q

Case 1 q>0  this creates repulsive force

Case 2 q<0 this creates attractive force

 If q is shifted from the centre then the positive charges nearer to this charge will attract it towards itself and charge q will never return to its centre.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- Gauss's law states that the total electric flux of an enclosed surface is given by q/? _0, where q is the charge enclosed by the surface.

So total charge inside the surface is = Q-2Q = -Q

Therefore total flux through the surface of the sphere = -Q/? ₀

Now, charge 5Q is lies outside the surface, thus it makes no contribution to electric flux through the given surface. So both option a and c are true.