Physics

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= $\frac{G{m}^2}{l^2}+\frac{G{m}^2}{\sqrt[]{3}{l}^2}+\frac{G{m}^2}{4l^2}=\frac{G{m}^2}{l^2}\left(1+\frac{1}{\sqrt[]{3}}+\frac{1}{4}\right)$

For numerical problems, start by finding the given data and know what needs to be calculated. To visualize the motion, it is advisable to use a diagram. Based on if the acceleration is uniform, choose the appropriate kinematic equations. Also, be careful with the unit conversions and sign conventions. Solve the numerical problems by following the step-by-step approach and at the end double-check your answer for accuracy.

Yes, if one practices from the NCERT Exemplar, they can score high in the CBSE Board exams and entrance exams like NEET and JEE. The exemplar questions include advanced and application-based questions which deepen students' understanding of key concepts. It contains information beyond the NCERT textbook and offers concept clarity to students. Practicing from the exemplar helps in improving the problem-solving skills of the students.

Motion in a Straight Line focuses on the rectilinear motion, which is the motion of objects along a single straight path. It covers key concepts such as speed, displacement, acceleration, velocity, and motion graphs. It helps students understand more complex types of motions.

If one solves all these questions then it is great but it is not something mandatory. Practicing these questions helps students develop a strong understanding of Coulomb's law, electric charges, the superposition principle, and electric field lines. Even practicing a significant portion helps in boosting exam confidence and helps students to score high in the examination.

The NCERT textbook is good to start with as it introduces students to the basic concepts and explains these topics to provide conceptual clarity. However, the exemplar focuses on reasoning, challenging multiple-choice questions and numerical problems. The exemplar is like the supplement that boosts students' grasp on these concepts given in the NCERT textbook through reasoning, multiple-choice and numerical problems.