physics ncert solutions class 11th

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

Explanation-

areal velocity of a planet revolving around the sun is constant with respect to time.

Explanation- examples of central force – gravitational force and electrostatic force

Examples of non central force = nuclear force, magnetic force acting between two current carrying loops etc.

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

Explanation-air molecules in the atmosphere are attracted vertically downward by the gravitational force of the earth just like an apple falling from the tree. Air molecules move randomly due to their thermal velocity and hence resultant motion of air molecules move randomly in vertical downward direction. But in case of apple it is heavier than air so it fall downwards only.

This is a Long Type Questions as classified in NCERT Exemplar

Explanation- r_p= radius of perihelion =2R

R_a = radius of aphelion =6R

So r_a=a(1+e)=6R

And r_p= a(1-e)=2R

Solving above eqns

 We get e = ½

By conservation of angular momentum angular momentum at perigee= angular momentum at apogee

So mv_pr_p=mv_ar_a

V_a/v_p=1/3

Where m is mass of satellite .

By conservation of energy , energy at perigee =energy at apogee

$\frac{1}{2}m{v}_p^2$ - $\frac{GMm}{r_p}=\frac{1}{2}m{v}_a^2-\frac{GMm}{r_a}$

So $v_p^2\left(1-\frac{1}{9}\right)=-2GM\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=2GM\left(\frac{1}{r_p}-\frac{1}{r_a}\right)$

V_p= $\frac{{\left[\frac{2GM}{R}\right(\frac{1}{2}-\frac{1}{6}\left)\right]}^{1/2}}{[{{[1-\frac{v_a}{v_p}]}^2]}^{1/2}}$ = $\sqrt[]{\frac{3}{4}\frac{GM}{R}}=6.85km/s$

V_p=6.85km/s , v_a=2.28km/s

So orbital velocity v_c= $\sqrt[]{\frac{GM}{r}}$

R=6R ,v_c= 3.23km/s

Hence to transfer to a circular orbit at apogee , we have to boost the velocity by 3.23-2.28=0.95km/s.

Explanation- mass of earth M = 6 $*{10}^{24}kg$

Radius of the earth , R = 6400km= 6.4 $*{10}^{6}m$

T=24h= 24 $*60*60=86400s$

G = 6.67 $*$ 10^-11Nm²/kg²

(a) Time period T = $2\pi \sqrt[]{\frac{{\left(R+h\right)}^3}{GM}}$

$T^2$ = $4\pi$ ^{2 $\frac{{(R+h)}^3}{GM}$}

$R+h$ = ( $\frac{T^{2}GM}{4{\pi }^2}$ )^1/3

$h=$  ( $\frac{T^{2}GM}{4{\pi }^2}$ )^1/3-R

So after solving we get h = 3.59 $*{10}^{7}m$

(b) If satellite is at height h from the earth's surface

Cos $\theta$ = $\frac{R}{R+h}=\frac{1}{\left(1+\frac{h}{R}\right)}=\frac{1}{1+5.61}=\frac{1}{6.61}=0.1513$ = cos81⁰18'

$\theta$ = 81⁰18'

$2\theta$ =2(81⁰18')= 162⁰36'

If n number of satellite needed to cover entire the earth then

So n = 360⁰/2 $\theta$ = 2.31

So minimum 3 satellite are required to cover entire earth.

This is a Long Type Questions as classified in NCERT Exemplar

Explanation- consider a diagram having vertices A,B,C,D,E and F

AC= AG+GC=2AG

= 2lcos30⁰= 2l $\frac{\sqrt[]{3}}{2}$

= $\sqrt[]{3}l=AE$

AD=AH+HJ+JD= lsin30⁰+l+lsin30⁰=2l

Force on mass m at A due to mass m at B is f₁= $\frac{Gmm}{l^2}$  along AB

Force on mass m at A due to mass m at C is f₂= $\frac{Gmm}{\sqrt[]{3}{l}^2}=\frac{G{m}^2}{3l^2}$  along AC

Force on mass m at A due to mass m at D is F₃= $\frac{Gmm}{({2l)}^2}=\frac{G{m}^2}{4l^2}$ along AD

Force on mass m at A due to mass m at E is F₄= $\frac{Gmm}{\sqrt[]{3}{l}^2}$ = $\frac{G{m}^2}{3l^2}$ along AE

Force on mass m at A due to mass m at F is F₅= $\frac{G{m}^2}{l^2}$  along AF

Resultant force due to F₁ and F₅ is F₁= $\sqrt[]{f_1^2+f_5^2+2f_1{f}_{2}cos60}$

= $\frac{\sqrt[]{3}G{m}^2}{3l^2}=\frac{G{m}^2}{\sqrt[]{3}{l}^2}$ along AD

So net force along AD = F₁+F₂+F₃= $\frac{G{m}^2}{l^2}+\frac{G{m}^2}{\sqrt[]{3}{l}^2}+\frac{G{m}^2}{4l^2}=\frac{G{m}^2}{l^2}\left(1+\frac{1}{\sqrt[]{3}}+\frac{1}{4}\right)$

4.32

(a) Let $\mathit{\theta }\left(\mathit{t}\right)=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{{\mathit{v}}_{\mathit{o}\mathit{y}}-\mathit{}\mathit{g}\mathit{t}}{{\mathit{v}}_{\mathit{o}\mathit{x}}}\right)$ be the angle at which the projectile is fired w.r.t. the x-axis, since ${\mathit{\theta }}_0=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{4{\mathit{h}}_{\mathit{m}}}{\mathit{R}}\right)$ depends on t

Therefore tan $\theta$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $\theta$ = $\theta \left(t\right)=\frac{Vx}{Vy}=\frac{Voy-gt}{Vox}$ sin2 $\theta \left(t\right)={\tan }^{-1}?\left(\frac{Voy-gt}{Vox}\right)$ /2g…….(1)

R = $h_{max}$ sin2 $u^2$ /g …….(2)

Dividing (1) by (2)

( $\theta$ /R) = [ $u^2$ sin2 $\theta$ /2g]/[ $h_{max}$ sin2 $u^2$ /g] = $\theta$ / 4

4.31

Speed of the cyclist = 27 km/h = 7.5 m/s

Radius of the road = 80m

The net acceleration is due to braking and the centripetal acceleration

Acceleration due to braking = 0.5 m/s²

Centripetal acceleration a = v²r = (7.5)2/80= 0.703 m/s²

The resultant acceleration is given by a = sqrt ( $a_t^2$ + $a_c^2$ ) = sqrt ( ${0.5}^2$ + ${0.7}^2$ ) = 0.86 m/s²

tan $\beta$ = AC / at = 0.7/0.5,  $\beta$ = 54.5 $°$