Physics Ncert Solutions Class 12th

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

Explanation-

(i) when r

$?E.ds$   = $\frac{1}{{\epsilon }_0}\int \mathrm{\rho }\mathrm{d}\mathrm{V}$ and we know that V = $\frac{4}{3}\pi r^3$

$dV=3*\frac{4}{3}\pi r^{3}dr$ = 4 $\pi r^{2}dr$

  $?E.ds$ = $\frac{1}{{\epsilon }_0}4\pi K\int r^{3}dr$

  E(4 ) $\pi r^2$ = $\frac{4\pi K{r}^4}{{\epsilon }_{0}4}$ = $\frac{k{r}^2}{4{\epsilon }_0}$ so it is clear that E is radially outwards.

  When r>R

$?E.ds$ = $\frac{1}{{\epsilon }_0}\int \mathrm{\rho }\mathrm{d}\mathrm{V}$

              E= $\frac{k{R}^4}{4{\epsilon }_0{r}^2}$ again field is outwards

(ii) When two protons are there then they must be on opposite sides or we can say along the end of diameter

So q= $\int \mathrm{\rho }\mathrm{d}\mathrm{V}=\int Kr4\pi r^{2}dr$

 .q= 4 $\pi K\frac{R^4}{4}=2e,soK=\frac{2e}{\pi R^4}$

If protons 1 and 2 are embedded at distance r from the center of the sphere then force will be

F=eE=- $\frac{eK{r}^2}{4{\epsilon }_0}$ but the force applied by proton F= $\frac{e^2}{4\pi {\epsilon }_0({2r)}^2}$

By adding these F= - $\frac{eK{r}^2}{4{\epsilon }_0}+\frac{e^2}{4\pi {\epsilon }_0({2r)}^2}$

If F=0 then $\frac{eK{r}^2}{4{\epsilon }_0}=\frac{e^2}{4\pi {\epsilon }_0({2r)}^2}$  so after solving r= $\frac{R}{8^{1/4}}$

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

Explanation- let us consider that universe is of radius R

             And we know that hydrogen is made up one electron and proton so net charge is

             -(1+y)e+e = -ye

             Now the number of hydrogen atom in the universe = N $*\frac{4}{3}\pi R^3$

             So total charge is = -ye $*$ N $*\frac{4}{3}\pi R^3$

             According to gauss theorem $?E.ds=\frac{q}{{\epsilon }_0}$

              So E $*4\pi R^2=\frac{-\mathrm{y}\mathrm{e}*\mathrm{}\mathrm{N}*\frac{4}{3}\pi R^3}{{\epsilon }_0}$

              E= -NRye/3 ${\epsilon }_0$

             And we know electrostatic force F=qE = -ye(-NRye/3 ) ${\epsilon }_0$ by using the value of E and q

             F= $\frac{y^2{e}^{2}NR}{3{\epsilon }_0}$ force is positive which shows that force is repulsive.

             But the gravity is g= $\frac{GM}{R^2}$ and gravitational force is F= $\frac{GMm}{R^2}$

             But M= N $*\frac{4}{3}\pi R^3{m}_p$  where $m_p$ is mass of proton

             So using this mass in above eqn so force = $\frac{G\mathrm{N}*\frac{4}{3}\pi R^3{m}_{p}m}{R^2}$

             But her masses of proton and hydrogen are equal so take as m

             F= $\frac{G\mathrm{N}*4\pi R{m}^2}{3}$

             So calculate its critical value electrostatic force and gravitational force both are equal.

             So, $\frac{G\mathrm{N}*4\pi R{m}^2}{3}=\frac{y^2{e}^{2}NR}{3{\epsilon }_0}$  

             So y²= 4 $\pi {\epsilon }_0\frac{m^2}{e^2}$  after using standard values we get value of y= approx. 10^-18

9.32 Focal length of the convex lens, $f_1$ = 30 cm

The liquid acts as a mirror, focal length of the liquid = $f_2$

Focal length of the system (convex lens + liquid), $f$ = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as

$\frac{1}{f}$ = $\frac{1}{f_1}$ + $\frac{1}{f_2}$ or $\frac{1}{f_2}=\frac{1}{f}-\frac{1}{f_1}$ = $\frac{1}{45}$ - $\frac{1}{30}$

$f_2=$ - 90 cm

Let the refractive index of the lens be ${\mu }_1$ and the radius of curvature of one surface be R

Hence, the radius of curvature of the other surface is –R

R can be obtained by using the relation

$\frac{1}{f_1}$ = ( ${\mu }_1-1\left)\right(\frac{1}{R}$ + $\frac{1}{\left|R\right|}$ ) = (1.5 – 1)( $\frac{2}{R})$

$\frac{1}{30}$ = $\frac{1}{R}$ , so R = 30 cm

Let the refractive index of the liquid be ${\mu }_2$

The radius of curvature of the liquid on the side of the plane mirror = $\infty$

Radius of curvature of the liquid on the side of the lens, R = -30 cm

The value of ${\mu }_2$ can be calculated using the relation,

$\frac{1}{f_2}$ = ( ${\mu }_2-1\left)\right(\frac{1}{-R}$ - $\frac{1}{\infty }$ )

$\frac{-1}{90}$ = ( ${\mu }_2-1\left)\right(\frac{1}{30}$ - 0)

${\mu }_2$ - 1 = $\frac{1}{3}$

${\mu }_2=\frac{4}{3}$ = 1.33

Hence, the refractive index of the liquid is 1.33.

9.30 Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of objective mirror,  $R_1$ = 220 mm

Hence focal length of the objective mirror,  $f_1$ = $\frac{R_1}{2}$ = 110 mm

Radius of curvature of secondary mirror,  $R_2$ = 140 mm

Hence focal length of the objective mirror,  $f_2$ = $\frac{R_2}{2}$ = 70 mm

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror. Hence, the virtual object distance for the secondary mirror,

u = $f_1-d$ = 110 – 20 = 90 mm

Applying the mirror formula for the secondary mirror, we can calculate the image distance ( $v$ ) as:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f_2}$ or $\frac{1}{v}$ = $\frac{1}{f_2}-\frac{1}{u}$ = $\frac{1}{70}$ - $\frac{1}{90}$

$v$ = 315 mm

Hence, the final image will be formed 315 mm away from the secondary mirror.

9.29 Focal length of the objective lens, $f_o$ = 140 cm

Focal length of the eyepiece, $f_e$ = 5 cm

Least distance of distinct vision, d = 25 cm

In normal adjustment, the separation between the objective lens and the eyepiece

= $f_o+f_e=140+5=145cm$

Height of the tower $h_1=100m$

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as : $\theta =\frac{h}{u}$ = $\frac{100}{3000}$ = $\frac{1}{30}$ rad

The angle subtended by the image produced by the objective lens is given as $\theta =\frac{h_2}{f_o}$ , where $h_2$ = height of the image of the tower formed by the objective lens

So, $h_2$ = $\theta *f_o$ = $\frac{1}{30}*140$ =4.7 cm

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation

m = 1 + $\frac{d}{f_e}$ = 1 + $\frac{25}{5}$ = 6

Height of the final image = m $*h_2$ = 6 $*4.7=28.2cm$