Physics

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

(a) consider a horizontal parcel of air with cross section A and height dh

Let the pressure on the top surface and bottom surface be P and p+dp. If the parcel is in equilibrium , then the net upward force must be balanced by the weight

(P+dP)-PA=- $\rho Adh*g$

dP= - $\rho gdh$

negative sign shows that pressure decreases with height.

(b) let $\rho$ _o be the density of air on the surface of earth.

As per question , pressure $\propto$  density

$\frac{P}{P_o}=\frac{\rho }{{\rho }_0}$

$\rho =\frac{{\rho }_o}{P_o}P$

dP= - $\frac{{\rho }_{o}g}{P_o}Pdh$

$\frac{dP}{p}=-\frac{{\rho }_{o}gdh}{P_o}$

${\int }_{P_o}^P\frac{dP}{P}=-\frac{{\rho }_{o}g}{P_o}{\int }_0^{h}dh$

In $\frac{P}{P_o}=-\frac{{\rho }_{o}gh}{P_o}$

P=P_oe(- $\frac{{\rho }_{o}gh}{P_o}$ )

(c) as P =P_{o $e^{\frac{-{\rho }_{o}gh}{P_o}}$}

in $\frac{P}{P_o}=-\frac{{\rho }_{o}gh}{P_o}$

p=1/10 P_o

in( $\frac{\frac{1}{10}{P}_o}{P_o}$ ) =- $\frac{-{\rho }_{o}gh}{P_o}$

in1/10 =- $\frac{{\rho }_{o}g}{P_o}h{\rho }_0$

h=- $\frac{P_o}{{\rho }_{o}g}$ in1/10= - $\frac{P_o}{P_{o}g}in\left(10\right)$ ^-1= $\frac{P_o}{P_{o}g}in\left(10\right)$

= $\frac{P_o}{P_{o}g}*2.303$

= $\frac{1.013*{10}^5}{1.22*9.8}*2.303=0.16*{10}^{5}m$

= 16 $*$ 10³m

(d) we know that

P $\propto \rho$  , temperature remains constant only near the surface of the earth , not at greater heights.

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