Let the planes earth be made of material of density p and is of radius R.

ஃ Mass of earth \(=M=\frac{4}{3} \pi R^{3} \rho\)
and acceleration due to gravity on earth = g
\(\therefore \ g=\frac{G M}{R^{2}}=\frac{G}{R^{2}} \times \frac{4}{3} \pi R^{3} \rho\)
\(\therefore \ g=G \cdot \frac{4}{3} \pi R \rho\)            ............(i)
At a depth d, the gravitational force is due to the mass distributed in the sphere of radius (R - d).
ஃ The acceleration due to gravity at the depth d
\(g^{\prime}=\frac{G M^{\prime}}{(R-d)^{2}}=G \frac{4}{3} \frac{\pi(R-d)^{3} \rho}{(R-d)^{2}}\)
\(g^{\prime}=G \frac{4}{3} \pi(R-d)\)        .............(ii)
Dividing g' by g, we get
\(\frac{g^{\prime}}{g}=\frac{R-d}{R}=1-\frac{d}{R}\)
\(\therefore \ g^{\prime}=g \frac{(R-d)}{R}=g\left(1-\frac{d}{R}\right)\)
ஃ g reduces as we move from surface to inwards and is zero at the centre.

Escape Speed: It is defined as the minimum speed with which the body has to be projected vertically upwards from the surface of earth (or any other planet) so that it just crosses the gravitational field of earth (or of that planet) and never returns on its own.
Let earth be a perfect sphere of mass M, radius R with centre at O. Let a body of mass m to be projected from a point A on the surface of earth, as shown in figure.
Join OA and produce it further.
P and Q are at a distance x and (x + dx) from the centre of the earth.
Gravitational force of attraction on the body at P is \(F=\frac{G M m}{x^{2}}\)
This much force has to be applied on the body to take the body in the upward direction. Work done in taking the body against gravitational attraction from P to Q is
\(d W=F d x=\frac{G M m}{x^{2}} d x\)
Total work done in taking the body against gravitational attraction from surface of earth (i.e. x = R) to a region beyond the gravitational field of earth (i.e. x = ∞) can be calculated by interchanging the above expression within the limits x = R to x = ∞. Thus, total work done is

\(W=\int_{R}^{\infty} \frac{G M m}{x^{2}} d x=G M m \int_{R}^{\infty} x^{-2} d x\)
\(=G M m\left[\frac{x^{-2+1}}{-2+1}\right]_{R}^{\infty}=-G M m\left[\frac{1}{x}\right]_{R}^{\infty}\)
\(=-G M m\left[\frac{1}{\infty}-\frac{1}{R}\right]=\frac{G M m}{R}\)
This work done is at the cost of kinetic energy given to the body at the surface of the earth.
K.E. of the body \(=\frac{1}{2} m v_{e}^{2}, v_{e}\) = escape speed of the body
\(\therefore \ \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}\)
or \(v_{e}^{2}=\frac{2 G M}{R} \text { or } v_{e}=\sqrt{\frac{2 G M}{R}}\)
or \(v_{e}=\sqrt{2 g R} \text { . }\) 

(i) Consider an object moving in a straight line with uniform acceleration' a', let at any instant of time 't', dx be the displacement of the objects.
\(\therefore\) instantaneous velocity \(v=\frac{d x}{d t} \text { or } d x=v \mathrm{dt}\)
dx = (u + at) dt (∵ v = u + at)
let x₀ and x be the displacements of the object at time 'zero' and 't'.
\(\int_{x_{0}}^{x} d x=\int_{0}^{t}(u+a t) d t=u \int_{0}^{t} d t+a \int_{0}^{t} t d t\)
\((x)_{x_{0}}^{x}=u(t)_{0}^{t}+a\left(\frac{t^{2}}{2}\right)_{0}^{t}\)
If x = x₀ = s (distance) covered by the object.
\(s=u t+\frac{1}{2} a t^{2}\)
(ii) Consider a particle moving in a straight line with initial velocity 'v' and acceleration 'a'
Then, \(a=\frac{d v}{d t}=\frac{d v}{d x} \times \frac{d x}{d t}=\frac{d v}{d x} \times v\)
adx = vdv
Integrating, \(\int_{x_{0}}^{x} a d x=\int_{u}^{V} v d v\)
\(\Rightarrow \quad a|x| x_{0}^{x}=\left.\frac{v^{2}}{2}\right|_{u} ^{V}\) 
\(a\left(x-x_{0}\right)=\frac{v^{2}}{2}-\frac{u^{2}}{2}\)
v² - u ² = 2a(x -x0 )
v ² - u ² = 2as 

Period of a revolution of a satellite is the time taken by the satellite to complete one revolution round the earth. It is denoted by T.
\(\therefore T=\frac { Circumference\ of\ circular\ orbit }{ Orbital\ velocity } \) 
or    \(T=\frac { 2\pi r }{ { v }_{ o } } \)
or     \(T=\frac { 2\pi (R+h) }{ { v }_{ o } } \quad \quad \quad \quad \quad [\therefore r=R+H]\)
or     \(T=2\pi (R+h)\sqrt { \frac { R+h }{ GM } } \left[ \because \quad { v }_{ o }=\sqrt { \frac { GM }{ R+h } } \right] \)
or     \(T=2\pi \sqrt { \frac { (R+h)^{ 2 } }{ GM } } \)     
Also,  \(T=2\pi \sqrt { \frac { (R+h)^{ 2 }(R+h) }{ GM } } \)
or       \(T=2\pi \sqrt { \frac { (R+h)^{ 3 } }{ gR^{ 2 } } } \)
 \(\because \quad \quad g{ R }^{ 2 }=GM\)  
\(\therefore T=2\pi \sqrt { \frac { (R+h)^{ 2 } }{ gR^{ 2 } } } \)   

Given \(V={ 2t }^{ 2 }-3t-10m/s\)
\({ a }_{ in }=\frac { dv }{ dt } =4t-3m/{ s }^{ 2 }\)
If t = 5 \({ a }_{ in }=5\times 4-3=17m/{ s }^{ 2 }\) 

The dimensions of LHS
= [M][LT^-2] = [ML²T^-2]
The dimensions of RHS = [M][LT^-1] = [ML²T^-2]
The dimensions of LHS and RHS are same and hence the consistency is verified.

The number of significant figures in the given quantities are given below.
(i) In 0.007, the number of significant figures is 1 because in a number less than 1, the zero's on the right of the decimal point but to the left of the first non-zero digit are not significant.
(ii) In 2.64 × 10²⁴ kg, the number of significant figures is 3 because all non-zero digits are significant, power of 10 are not taken in significant figure.
(iii) In 0.2370, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant.
(iv) In 6.320, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant.
(v) In 6.032, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant.
(vi) In 0.0006032, the number of significant figures is 4, because in a number less than 1, the zero's on the right of the decimal point but to the left of the first non-zero digit are not significant.

As,      \(x=\frac { { a }^{ 3 }{ b }^{ 3 } }{ c\sqrt { d } } \) 
 \(\therefore \quad\frac { \triangle x }{ x } =\pm \left[ 2\frac { \triangle a }{ a } +3\frac { \triangle b }{ b } +\frac { \triangle c }{ c } +\frac { 1 }{ 2 } \frac { \triangle d }{ d } \right] \)     
Percentage error in x is given by 
 \(\frac { \triangle x }{ x } \times 100\) = \(\pm\) \([2\times 2\)% + \(3\times 1\)% +3% + \(\frac { 1 }{ 2 } \times 4\)%\(]\) 
 \(=\pm 12\)%

\(g^{\prime}=g\left(1-\frac{d}{R}\right)\)
\(=9.8\left(1-\frac{30 \times 1000}{6.3 \times(1000)^{2}}\right)\)
\(=9.8\left(1-\frac{1}{210}\right)\)
\(=9.8\left(\frac{209}{210}\right)=9.75 \mathrm{~ms}^{-2}\) 

(i) The planets including earth, go around the sun in elliptical orbits.
(ii) The line joining the Sun and the planet sweeps equal areas in equal intervals of time.
(iii) The square of the time period of revolution is directly proportional to the cube of the semi-major axis of the elliptical orbit.
Since \(T^{2} \propto r^{3}\) we have,
\(\left(\frac{2 \pi r}{v}\right)^{2} \propto r^{3}\)
\(v^{2}=4 \pi^{2} \frac{r^{2}}{r^{3}}=\frac{4 \pi^{2}}{r}\)
\(\frac{m v^{2}}{r}=\frac{4 m \pi^{2}}{r^{2}}\)
The centripetal force \(\frac{m v^{2}}{r}\) is caused by M - earth on the planet of mass m.
Thus \(\mathrm{F} \propto \frac{\mathrm{Mm}}{r^{2}}\)
It is the Newton's Universal Law of Gravitation.

Velocity after 6 seconds = 10 + 2 x 6 = 22 ms^-1 .

Length of actual path covered between the initial and final points is distance while the length of the shortest path between initial and final points is displacement. The magnitude of displacement can be both positive and negative while distance is always positive.

\(KE=\frac { 1 }{ 2 } mv^{ 2 }\text{ i.e.,dimensional formula of KE is} [ML^{ 2 }T^{ -2 }]\) 
\(Pressure=\frac { Force }{ Area } =\frac { [MLT^{ -2 }] }{ [L^{ 2 }] } =[ML^{ -1 }T^{ -2 }]\)

0.001293 g/cc

During acceleration,
vm = 0 + 5 x t_a
\(\Rightarrow \quad t_{a}=\frac{v_{m}}{5}\)
\(v_{m}^{2}=0+2 \times 5 \times s_{a}\)
\(\Rightarrow \quad s_{a}=\frac{v_{m}^{2}}{10}\)

During deceleration,
\(0=v_{m}-3 t_{d} \Rightarrow t_{d}=\frac{v_{m}}{3}\)
\(0=v_{m}^{2}-2 \times 3 \times s_{d} \Rightarrow s_{d}=\frac{v_{m}^{2}}{6}\)
Total time = 6 = ta + td.
\(\therefore \quad v_{\mathrm{m}}=\frac{6 \times 5 \times 3}{8}=11.25 \mathrm{~ms}^{-1}\)
Total length covered \(=s_{a}+s_{d}=v_{m}^{2}\left(\frac{1}{10}+\frac{1}{6}\right)\)
\(=(11.25)^{2} \frac{16}{60}=33.75 \mathrm{~m}\) 

(i) Velocity attained by a particle after time t:
Let dt: be the change in velocity of the particle in time dt. Therefore, the acceleration of the particle is given by
\(a=\frac{dv}{dt} \ or \ dv=a \ dt\)
By integrating both sides, we get
\(\int{dv}= \int{a dt}\)
or \(\int{dv}= a \int{ dt}\)
or    v = at + k   --- (i)
where k is constant of integration.
when t = 0, v = u
Putting these values in equation (i), we get
k = u
Now putting the value of k in equation (i), we get
v = u + at
(ii) Displacement of the particle after time t:
Let dx be the displacement of the particle in time dt. Therefore, the velocity of the particle is given by
\(v=\frac{dx}{dt}\ or \ dx=vdt\)
Since v  =u + at
∴ dx=(u+at)dt
Integrating both sides, we get
\(\int{dx}=\int{(u+at)}dt\)
or  \(\int{dx}=\int{u}dt+\int{at \ dt}\) 
\(x= u\int{dt}+a \int{t\ dt}\)     [∵ u and a are constants]
or \(x=ut+a \frac{t^{2}}{2}+k\)
where k is constant of proportionality
where t = 0, x = x₀
∴ from equation (ii), we get
\(x=x_{0}+ut+\frac{1}{2}at^{2}\)
or \(x-x_{0}=ut+\frac{1}{2}at^{2}\)
since  x-x₀= S, displacement of the particle in the time interval t.
S = \(ut+\frac{1}{2}at^{2}\)
(iii) Velocity attained by a particle after travelling a distance S:
We know, \(v=\frac{dx}{dt}\)
Multiplying and dividing R.H.S. by dv, we get
\(v=\frac{dx}{dt}.\frac{dv}{dv}=\frac{dx}{dv}.\frac{dv}{dt}\)
As \(\frac{dv}{dt}=a (acceleration)\)
∴ v = a\(\frac { dx }{ dv } \) or v dv=a dx
Integrating both sides, we get \(\int { v\ dv=\int { a\ dx=a\int { dx } } } \)
or = \(\frac { v^{ 2 } }{ 2 } \)ax+k
when x = 0,v = u
Then,from eqn.(i),k = \(\frac { u^{ 2 } }{ 2 } \)
Putting the value of k in eqn. (i), we get
\(\frac { v^{ 2 } }{ 2 } \)-ax+\(\frac { u^{ 2 } }{ 2 } \)
or \(\frac { v^{ 2 } }{ 2 } -\frac { u^{ 2 } }{ 2 } =ax\)
or v²-u²= 2ax
x = s, then
v²-u² = 2 aS.

For Kepler's laws of planetary motion, please see facts that matter.
The body will become weightless if the gravitational force mg on it is entirely used up in providing the centripetal acceleration for the rotation of the earth,
Then \(mg=\frac{m\theta^2}{R}=m\omega^2R\)
\(\omega^2=\frac{g}{R}=\frac{9.8}{6400\times10^3}\)
ω =1.237 x 10^-3 rad s^-1
If the earth rotates at this speed, the bodies on the equator will have no weight. At a latitude ф the apparent weight W_A is given by
\(W_A=mg(1-\frac{\omega^2R}{g}cos^2\phi)\)
Here, however, g = ω²R
Therefore,​​​​​​​ W_A= mg(1-cos² ф )
When  ф ​​​​​​​= 60⁰,cos ф ​​​​​​​= \(\frac{1}{2}\)
\(W_A=mg(1-\frac{1}{4})=\frac{3}{4}\times true\ weight\)
At poles, \(\phi=\frac{\pi}{2}\)
W_A mg = true weight
Thus, a body situated on the poles remains unaffected, whatever the speed of rotation of the earth.​​​​​​​

\(2.7972\times { 10 }^{ -7 }N\)

Let the frequency of vibration of the string be given by 
\(\nu =Kl^{ a }m^{ b }T^{ c }\) ......(i)
where K = a dimensionless constant
Dimensions of the various quantities are
\(\nu =\left[ T^{ -1 } \right] ,l=\left[ L \right] ,T=\left[ T \right] \)
Force =\(\left[ MLT^{ -2 } \right] \)
and \(m=\frac { mass }{ length } =\left[ ML^{ -1 } \right] \)
Substituting these dimensions in equation(i), we get
\(\left[ T^{ -1 } \right] ={ \left[ L \right] }^{ a }\left[ ML^{ -1 } \right] ^{ b }\left[ MLT^{ -2 } \right] ^{ c }\)
or \(\left[ M^{ 0 }L^{ 0 }T^{ -1 } \right] =\left[ { M }^{ b+c }{ L }^{ a-b+c }{ T }^{ -2c } \right] \)
Equating the dimensions of M,L and T, we get
b + c = 0, a - b + c=0 and - 2c = - 1
on solving,\(a=-1,b=-\frac { 1 }{ 2 } and\quad c=\frac { 1 }{ 2 } \)
\(\therefore \quad \left( \nu \right) =Kl^{ -1 }m^{ { -1 }/{ 2 } }T^{ { 1 }/{ 2 } }\quad or\quad \left( \nu \right) =\frac { K }{ l } \sqrt { \frac { T }{ m } } \) 

According to Newton's law gravitation, the force F, between two masses m₁ and m₂ separated by distance r can be given as
 \(F=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \)
\(\\ Where,\ G=universal\ gravitational\ constant\)
\(G=\frac { Fr^{ 2 } }{ { m }_{ 1 }{ m }_{ 2 } } \)
\(\\ G=\frac { Newton\times (metre)^{ 2 } }{ (kg)^{ 2 } } \)
\(\\ G=\frac { (mass\times acceleration)\times (metre)^{ 2 } }{ (mass)^{ 2 } } \)
\(\\ =\frac { 1 }{ mass } \left( \frac { Change\quad in\quad velocity }{ Time } \right) \times (Length)^{ 2 }\)
\(\\ G=\frac { (Length)^{ 2 } }{ Mass\times Time } \times \frac { Distance }{ Time } \)
\(\\ G=\frac { [L]^{ 2 } }{ [M]\times [T] } \times \frac { [T] }{ [T] } =[{ M }^{ -1 }{ l }^{ 3 }{ T }^{ -2 }]\)

D) If the assertion and reason both are false.

A) If both assertion and reason are true and the reason is the correct explanation of the assertion.
Unit of quantity (L/R) is Henry/ohm. As Henry = ohm ´ sec, hence unit of L/R is sec i.e. [L/R] = [T]. Similarly, unit of product CR is farad ´ ohm or, \(\frac{\text { Coulomb }}{\text { Volt }} \times \frac{\text { Volt }}{A m p} \mid \text { or, } \frac{S e c \times A m p}{A m p} \mid \text {, }\) or, sec i.e. [CR] = [T] therefore [L/R] and [CR] both have the same dimension.

(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion

D) If the assertion and reason both are false.
Light year measures distance and year measures time. One light year is the distance traveled by light in one year.