Newton’s Law of Gravitation Questions in English

(B) No,the inertial mass and gravitational mass of a body are equivalent to each other. According to the Equivalence Principle in general relativity,the inertial mass (which determines the resistance to acceleration) and the gravitational mass (which determines the gravitational force experienced in a field) are identical for all objects.

(N/A) The value of the gravitational constant $G$ can be determined experimentally,which was first done by Cavendish in $1798$. The apparatus used is schematically shown in the figure.
The bar $AB$ has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.
Two large lead spheres are brought close to the small ones but on opposite sides. The big spheres attract the nearby small ones by equal and opposite forces. There is no net force on the bar,but there is a torque equal to $F \times L$,where $F$ is the force of attraction between a big sphere and a small sphere,and $L$ is the length of the bar. As a result,the bar rotates about the wire $OM$. The wire gets twisted until the restoring torque of the wire equals the gravitational torque.
In this equilibrium position,the angle of twist $\theta$ is measured. If $k$ is the torsion constant of the wire,the restoring torque is $\tau = k\theta$.
Gravitational force on each small sphere is $F = \frac{GMm}{d^2}$,where $M$ is the mass of the large sphere,$m$ is the mass of the small sphere,and $d$ is the distance between their centers.
The gravitational torque is $\tau_g = F \times L = \frac{GMm}{d^2} L$.
At equilibrium,$\tau_g = \tau$,so $\frac{GMm}{d^2} L = k\theta$.
Thus,$G = \frac{k\theta d^2}{MLm}$.

(A) The gravitational force $F$ between two bodies of masses $M$ and $m$ separated by a distance $r$ is given by Newton's Law of Gravitation: $F = G \frac{Mm}{r^2}$.
As the distance $r$ approaches infinity $(r \to \infty)$,the denominator $r^2$ becomes infinitely large.
Therefore,the force $F = G \frac{Mm}{r^2}$ approaches zero.
Thus,at an infinite distance from the Earth,the gravitational force acting on a body is zero.

(N/A) According to Newton's law of gravitation,the force between two masses is given by $F = \frac{G m_{1} m_{2}}{r^{2}}$.
If we take $m_{1} = 1 \ kg$,$m_{2} = 1 \ kg$,and $r = 1 \ m$,then $F = G$.
Therefore,the universal gravitational constant is defined as the gravitational force of attraction between two bodies of unit mass separated by a unit distance.

(A) $G$ is the Universal Gravitational Constant.
Its value remains the same at any place and at any time in the universe.
Therefore,the value of $G$ on the Moon is also $6.67 \times 10^{-11} \ Nm^2/kg^2$.

(B) The gravitational force between two objects is given by $F = G \frac{m_1 m_2}{r^2}$.
This formula shows that the gravitational force depends only on the masses of the objects and the distance between them.
It does not depend on the medium in which the objects are placed.
Therefore,the gravitational force in air and water remains the same.
Thus,the ratio of the force in air to the force in water is $1:1$.

(N/A) The mass $m$ of each sphere can be expressed in terms of density $\rho$ and radius $r$ as $m = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 \rho$.
The distance between the centers of the two spheres in contact is $d = r + r = 2r$.
According to Newton's Law of Gravitation,the force $F$ is given by $F = \frac{G m_1 m_2}{d^2}$.
Substituting the values,$F = \frac{G (\frac{4}{3} \pi r^3 \rho) (\frac{4}{3} \pi r^3 \rho)}{(2r)^2}$.
$F = \frac{G \cdot \frac{16}{9} \pi^2 r^6 \rho^2}{4r^2}$.
$F = (\frac{4}{9} \pi^2 \rho^2 G) r^4$.
Since $\frac{4}{9} \pi^2 \rho^2 G$ is a constant,we have $F \propto r^4$.

(B) According to Newton's Law of Universal Gravitation,the force between two bodies is given by $F = G \frac{m_1 m_2}{r^2}$.
This force depends on the product of the masses of the two bodies $(m_1 m_2)$.
Since the force is mutual,the Earth exerts the same magnitude of force on the Moon as the Moon exerts on the Earth,in accordance with Newton's Third Law of Motion.
Therefore,the gravitational force exerted by the Moon on the Earth is equal to the gravitational force exerted by the Earth on the Moon.

(B) According to Newton's second law of motion,$F = ma$,or $a = F/m$.
Since the force $F$ exerted by the Earth on the apple and the apple on the Earth is equal in magnitude,the acceleration produced is inversely proportional to the mass of the object.
The mass of the Earth $(M)$ is extremely large compared to the mass of the apple $(m)$.
Therefore,the acceleration of the Earth $(a_E = F/M)$ is so small that it is practically undetectable,whereas the acceleration of the apple $(a_a = F/m)$ is significant,causing it to fall towards the Earth.

(C) According to Newton's law of universal gravitation,the force $F$ is given by $F = G \frac{m_1 m_2}{r^2}$.
Since the gravitational constant $G$ is extremely small $(6.67 \times 10^{-11} \ N \ m^2/kg^2)$,the force $F$ between two ordinary objects with small masses $m_1$ and $m_2$ is negligible.
Because this force is so weak,it is insufficient to overcome other forces like friction or air resistance,and it is not strong enough to cause observable acceleration towards each other.

(A) According to Newton's law of universal gravitation,the force $F$ between two objects of masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
This implies that $F \propto \frac{1}{r^2}$.
Let the initial force be $F_1 = 1 \ N$ at distance $r_1 = r$.
When the distance is doubled,$r_2 = 2r$.
The new force $F_2$ is given by the ratio: $\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the values: $\frac{F_2}{1} = \left(\frac{r}{2r}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Therefore,$F_2 = \frac{1}{4} \ N = 0.25 \ N$.

(A) The formula for the gravitational force between two objects is $F = \frac{G m_1 m_2}{r^2}$.
Given: $m_1 = 1\,kg$,$m_2 = 1\,kg$,$r = 10\,cm = 0.1\,m = 10^{-1}\,m$,and $G = 6.67 \times 10^{-11}\,N\cdot m^2/kg^2$.
Substituting the values into the formula:
$F = \frac{(6.67 \times 10^{-11}) \times (1) \times (1)}{(10^{-1})^2}$
$F = \frac{6.67 \times 10^{-11}}{10^{-2}}$
$F = 6.67 \times 10^{-9}\,N$.

(B) According to Newton's law of universal gravitation,the force exerted by the Sun on a planet is given by $F = \frac{G M m}{r^2}$,where $M$ is the mass of the Sun,$m$ is the mass of the planet,and $r$ is the distance between them.
Let $m_e$ and $m_p$ be the masses of Earth and Pluto,and $r_e$ and $r_p$ be their respective distances from the Sun.
Given $m_e = m_p$ and $r_p = 40 r_e$.
The gravitational force on Earth is $F_e = \frac{G M m_e}{r_e^2}$.
The gravitational force on Pluto is $F_p = \frac{G M m_p}{r_p^2}$.
The ratio of the forces is $\frac{F_e}{F_p} = \frac{G M m_e / r_e^2}{G M m_p / r_p^2} = \frac{r_p^2}{r_e^2}$.
Substituting $r_p = 40 r_e$,we get $\frac{F_e}{F_p} = \frac{(40 r_e)^2}{r_e^2} = 40^2 = 1600$.
Thus,the ratio is $1600: 1$.

(A) According to the inverse square law,the gravitational force between $A$ and $B$ is given by:
$F = G \frac{m_A m_B}{r^2}$
The acceleration of object $A$ is $a_A = \frac{F}{m_A} = \frac{G m_B}{r^2}$ ... $(1)$
For the second case,where the force is inversely proportional to the fourth power of the distance:
$F' = G \frac{m_A m_B}{r^4}$
The new acceleration of object $A$ is $a_A' = \frac{F'}{m_A} = \frac{G m_B}{r^4}$
Substituting the expression from equation $(1)$ into the new acceleration formula:
$a_A' = \frac{1}{r^2} \left( \frac{G m_B}{r^2} \right) = \frac{a_A}{r^2}$
Therefore,the acceleration of object $A$ becomes $\frac{a_A}{r^2}$.

(C) The gravitational force between two point masses is given by Newton's law of gravitation: $F = \frac{G m_1 m_2}{r^2}$.
This formula shows that the gravitational force depends only on the masses $(m_1, m_2)$ and the distance $(r)$ between them.
It does not depend on the medium present between the masses.
Therefore,when the masses are dipped in water while keeping the separation constant,the gravitational force remains unchanged.

(N/A) No,we cannot shield a body from gravitational influence.
Gravitational force is a fundamental interaction that acts between any two masses in the universe.
Unlike electric fields,which can be shielded by placing a charge inside a conductor (where the internal field becomes zero due to charge redistribution),gravitational fields cannot be shielded.
There is no such thing as a 'gravitational insulator' or a material that can block gravitational flux.
Gravitational force depends only on the masses of the two bodies and the distance between them,as given by Newton's Law of Gravitation: $F = G \frac{m_1 m_2}{r^2}$.
Therefore,placing a body inside a hollow sphere does not eliminate the gravitational pull exerted by external matter.

(N/A) According to the shell theorem for gravitation:
$1$. For a point mass $m$ located outside the hollow spherical shell $(r \geq R)$,the shell acts as if all its mass $M$ is concentrated at its centre. Thus,the gravitational force is given by $F = \frac{GMm}{r^2}$.
$2$. For a point mass $m$ located inside the hollow spherical shell $(r < R)$,the net gravitational force exerted by the shell on the point mass is zero,because the gravitational fields from different parts of the shell cancel each other out. Thus,$F = 0$.
The graph of $F$ versus $r$ is zero for $0 \leq r < R$ and follows an inverse-square law $(F \propto 1/r^2)$ for $r \geq R$.

(N/A) Let the vertices of the regular hexagon be $A, B, C, D, E, F$ in order. Consider the mass at vertex $A$. The forces acting on it due to the other five masses are:
$1$. Force due to mass at $B$: $F_1 = \frac{Gm^2}{l^2}$ along $\vec{AB}$.
$2$. Force due to mass at $F$: $F_5 = \frac{Gm^2}{l^2}$ along $\vec{AF}$.
$3$. Force due to mass at $C$: $F_2 = \frac{Gm^2}{(AC)^2} = \frac{Gm^2}{3l^2}$ along $\vec{AC}$.
$4$. Force due to mass at $E$: $F_4 = \frac{Gm^2}{(AE)^2} = \frac{Gm^2}{3l^2}$ along $\vec{AE}$.
$5$. Force due to mass at $D$: $F_3 = \frac{Gm^2}{(AD)^2} = \frac{Gm^2}{4l^2}$ along $\vec{AD}$.
By symmetry,the components of $F_1$ and $F_5$ perpendicular to the diagonal $AD$ cancel out. Their components along $AD$ are $F_1 \cos 30^{\circ} + F_5 \cos 30^{\circ} = 2 \cdot \frac{Gm^2}{l^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}Gm^2}{l^2}$.
Similarly,the components of $F_2$ and $F_4$ perpendicular to $AD$ cancel out. Their components along $AD$ are $F_2 \cos 30^{\circ} + F_4 \cos 30^{\circ} = 2 \cdot \frac{Gm^2}{3l^2} \cdot \frac{\sqrt{3}}{2} = \frac{Gm^2}{\sqrt{3}l^2}$.
The force $F_3$ is already along $AD$.
Total force $F_{net} = \frac{\sqrt{3}Gm^2}{l^2} + \frac{Gm^2}{\sqrt{3}l^2} + \frac{Gm^2}{4l^2} = \frac{Gm^2}{l^2} (\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{4}) = \frac{Gm^2}{l^2} (\frac{3+1}{\sqrt{3}} + \frac{1}{4}) = \frac{Gm^2}{l^2} (\frac{4}{\sqrt{3}} + \frac{1}{4}) = \frac{Gm^2}{l^2} (\frac{16+\sqrt{3}}{4\sqrt{3}})$.

(C) According to Newton's law of gravitation,the initial force between two objects of mass $m$ separated by distance $r$ is given by $F = \frac{Gm^2}{r^2}$.
When one-third of the mass of one object is transferred to the other,the new masses become $m_1 = m - \frac{1}{3}m = \frac{2}{3}m$ and $m_2 = m + \frac{1}{3}m = \frac{4}{3}m$.
The new gravitational force $F^{\prime}$ is given by $F^{\prime} = \frac{G m_1 m_2}{r^2}$.
Substituting the new masses: $F^{\prime} = \frac{G (\frac{2}{3}m) (\frac{4}{3}m)}{r^2} = \frac{8}{9} \frac{Gm^2}{r^2}$.
Since $F = \frac{Gm^2}{r^2}$,we get $F^{\prime} = \frac{8}{9}F$.

(C) The initial gravitational force is given by $F = \frac{GMm}{R^2}$.
When the masses are tripled $(M' = 3M, m' = 3m)$ and the distance is doubled $(R' = 2R)$,the new force $F'$ is:
$F' = \frac{G(3M)(3m)}{(2R)^2} = \frac{9GMm}{4R^2} = 2.25F$.
The change in force is $\Delta F = F' - F = 2.25F - F = 1.25F$.
The percentage increase is $\frac{\Delta F}{F} \times 100\% = 1.25 \times 100\% = 125\%$.

(C) Let $y$ be the amount of mass transferred from the object with mass $3x$ to the object with mass $2x$.
After the transfer,the new masses are $(3x - y)$ and $(2x + y)$.
The gravitational force between them is given by $F' = \frac{G(3x - y)(2x + y)}{r^2}$.
To maximize the force,we differentiate $F'$ with respect to $y$ and set it to zero:
$\frac{dF'}{dy} = \frac{G}{r^2} \frac{d}{dy} [6x^2 + 3xy - 2xy - y^2] = 0$.
$\frac{G}{r^2} [3x - 2x - 2y] = 0$.
$x - 2y = 0$.
$y = \frac{x}{2}$.
Thus,the mass to be transferred is $\frac{x}{2}$.

(A) The gravitational force on the point mass $m$ due to the inner uniform sphere of mass $M$ is given by Newton's law of gravitation as $F_1 = \frac{G M m}{x^2}$,acting towards the center.
The gravitational force on the point mass $m$ due to the outer spherical shell of mass $M$ and radius $2R$ is zero. This is because,according to the shell theorem,the gravitational field inside a uniform spherical shell is zero everywhere,as the gravitational forces exerted by different parts of the shell on the point mass cancel each other out completely.
Therefore,the net gravitational force on the particle is $F_{net} = F_1 + 0 = \frac{G M m}{x^2}$.

(D) The correct option is $D$.
The gravitational constant,denoted by $G$,is a universal constant.
It is independent of the size of the bodies,their gravitational mass,the distance between them,or the medium in which they are placed.
Therefore,it does not depend on any of the factors mentioned in options $A$,$B$,or $C$.

(B) The phenomenon of gravitation is a universal force of attraction that acts between any two bodies in the universe,regardless of their shape,size,or mass.
While Newton's law of gravitation is strictly derived for point masses,it can be extended to bodies of any arbitrary shape by considering them as a collection of point masses and integrating the gravitational forces acting on each infinitesimal element.
Therefore,gravitation is the phenomenon of interaction between any arbitrary shaped masses.

(C) Let the third point mass $m_0$ be placed at a distance $r$ from the mass $m$. Then its distance from the mass $4m$ is $(d-r)$.
The gravitational force on $m_0$ due to mass $m$ is $F_1 = \frac{G m m_0}{r^2}$.
The gravitational force on $m_0$ due to mass $4m$ is $F_2 = \frac{G (4m) m_0}{(d-r)^2}$.
For the net gravitational force on $m_0$ to be zero,the magnitudes of these forces must be equal:
$F_1 = F_2$
$\frac{G m m_0}{r^2} = \frac{4 G m m_0}{(d-r)^2}$
$\frac{1}{r^2} = \frac{4}{(d-r)^2}$
Taking the square root on both sides:
$\frac{1}{r} = \frac{2}{d-r}$
$d - r = 2r$
$d = 3r$
$r = \frac{d}{3}$
Therefore,the distance of the point from the mass $m$ is $\frac{d}{3}$.

(B) Let $F_1, F_2, F_4, F_8, \dots$ be the gravitational forces exerted on the mass at the origin by the masses $m$ located at $x = 1, 2, 4, 8, \dots$ respectively.
Using Newton's law of gravitation,$F = \frac{G m_1 m_2}{r^2}$,the forces are:
$F_1 = \frac{G m^2}{1^2} = G m^2$
$F_2 = \frac{G m^2}{2^2} = \frac{G m^2}{4}$
$F_4 = \frac{G m^2}{4^2} = \frac{G m^2}{16}$
$F_8 = \frac{G m^2}{8^2} = \frac{G m^2}{64}$
The total force $F_{total}$ is the sum of these forces:
$F_{total} = G m^2 \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \right)$
The term in the bracket is an infinite geometric progression ($G$.$P$.) with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.
The sum of an infinite $G$.$P$. is given by $S = \frac{a}{1 - r}$.
$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Therefore,$F_{total} = G m^2 \left( \frac{4}{3} \right) = \frac{4}{3} G m^2$.

(D) The gravitational force exerted by particle $B$ on particle $A$ is $F = \frac{G m^2}{L^2}$ directed towards $B$.
Similarly,the gravitational force exerted by particle $C$ on particle $A$ is $F = \frac{G m^2}{L^2}$ directed towards $C$.
The angle between these two forces is $60^{\circ}$ because the triangle is equilateral.
The resultant force $F_{\text{net}}$ on particle $A$ is given by the vector sum:
$F_{\text{net}} = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^{\circ}} = \sqrt{2 F^2 + 2 F^2 (0.5)} = \sqrt{3 F^2} = \sqrt{3} F$
Substituting $F = \frac{G m^2}{L^2}$,we get $F_{\text{net}} = \sqrt{3} \frac{G m^2}{L^2}$.
According to Newton's second law,the acceleration $a$ of particle $A$ is $a = \frac{F_{\text{net}}}{m}$.
$a = \frac{\sqrt{3} G m^2}{L^2 m} = \sqrt{3} \frac{G m}{L^2}$.

(A) The gravitational force is a conservative force.
By definition,the work done by a conservative force in moving an object between two points is independent of the path taken.
It depends only on the initial and final positions of the object.
Since all three paths start at point $A$ and end at point $B$,the work done by the gravitational force along each path must be equal.
Therefore,$W_1 = W_2 = W_3$.

(C) The gravitational force between the two particles acts as the centripetal force required for circular motion.
The distance between the two particles is $r = 2a$.
The gravitational force is given by $F = \frac{G m m}{(2a)^2} = \frac{G m^2}{4a^2}$.
The centripetal force required for a particle of mass $m$ to move in a circle of radius $a$ with angular speed $\omega$ is $F_c = m \omega^2 a$.
Equating the two forces: $\frac{G m^2}{4a^2} = m \omega^2 a$.
$\omega^2 = \frac{G m}{4 a^3}$.
Therefore,the angular speed is $\omega = \sqrt{\frac{G m}{4 a^3}}$.

(B) Let the side length of the square be $a$. Consider one of the masses at a corner. It experiences gravitational forces from the other three masses.
Two masses are at a distance $a$ (adjacent corners),and one mass is at a distance $\sqrt{2}a$ (opposite corner).
The forces from the two adjacent masses are $F = \frac{Gm^2}{a^2}$ each,acting at an angle of $90^\circ$. Their resultant is $\sqrt{F^2 + F^2} = \sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$.
The force from the opposite mass is $F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$.
Both these forces act along the diagonal of the square,so the net force is:
$F_{\text{net}} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$.
Given that $F_{\text{net}} = \left( \frac{2\sqrt{2} + 1}{32} \right) \frac{Gm^2}{L^2}$,we equate the two expressions:
$\frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right) = \left( \frac{2\sqrt{2} + 1}{32} \right) \frac{Gm^2}{L^2}$.
$\frac{1}{2a^2} = \frac{1}{32L^2} \implies a^2 = 16L^2 \implies a = 4L$.

(D) Let the vertices of the equilateral triangle be $P, Q,$ and $R$. The mass $m_2$ is placed at point $S$,which is the midpoint of side $QR$.
The gravitational forces exerted by the masses at $Q$ and $R$ on the mass $m_2$ at $S$ are equal in magnitude and opposite in direction because $S$ is the midpoint of $QR$. Thus,these two forces cancel each other out.
The net force on $m_2$ is only due to the mass $m_1$ located at vertex $P$.
In $\triangle PQS$,the side $PQ = \frac{L}{3}$ and $\angle PQS = 60^{\circ}$. The distance $h$ (the altitude from $P$ to $QR$) is given by $h = PQ \sin 60^{\circ} = \frac{L}{3} \cdot \frac{\sqrt{3}}{2} = \frac{L\sqrt{3}}{6}$.
The gravitational force $F$ on $m_2$ due to $m_1$ at $P$ is:
$F = \frac{G m_1 m_2}{h^2} = \frac{G m_1 m_2}{(\frac{L\sqrt{3}}{6})^2} = \frac{G m_1 m_2}{\frac{3L^2}{36}} = \frac{G m_1 m_2}{\frac{L^2}{12}} = \frac{12 G m_1 m_2}{L^2}$.

(B) The gravitational force between two spheres of mass $M$ separated by a distance $d$ is given by $F = \frac{GM^2}{d^2}$.
Since the spheres are in contact,the distance between their centers is $d = 2R$.
Thus,$F = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}$.
The mass $M$ of each sphere is given by $M = \text{density} \times \text{volume} = \rho \times \frac{4}{3} \pi R^3$.
Substituting the value of $M$ into the force equation:
$F = \frac{G}{4R^2} \left( \rho \cdot \frac{4}{3} \pi R^3 \right)^2$
$F = \frac{G}{4R^2} \cdot \rho^2 \cdot \frac{16}{9} \pi^2 R^6$
$F = \frac{4}{9} G \pi^2 R^4 \rho^2$
Therefore,$F \propto R^4 \rho^2$.

(B) The gravitational force $F$ between two bodies of mass $m$ separated by a distance $d$ is given by $F = \frac{G m^2}{d^2}$.
Since the spheres are in contact,the distance between their centers is $d = R + R = 2R$.
The mass $m$ of each sphere with density $\rho$ is $m = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$.
Substituting these into the force equation:
$F = \frac{G (\frac{4}{3} \pi R^3 \rho)^2}{(2R)^2}$
$F = \frac{G \times \frac{16}{9} \pi^2 R^6 \rho^2}{4 R^2}$
$F = \frac{4}{9} G \pi^2 \rho^2 R^4$
Since $G$,$\pi$,and $\rho$ are constants,$F \propto R^4$.
Therefore,$x = 4$.

(A) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the mass $m_{2}$ be placed at point $D$,which is the midpoint of side $BC$.
The gravitational forces on $m_{2}$ due to the masses $m_{1}$ at $B$ and $C$ are equal in magnitude and opposite in direction,so they cancel each other out.
The distance of point $D$ from vertex $A$ is the altitude $h$ of the equilateral triangle.
In the right-angled triangle $ABD$,the hypotenuse $AB = \frac{L}{3}$ and $\angle BAD = 30^{\circ}$.
Thus,$h = AB \cos 30^{\circ} = \frac{L}{3} \times \frac{\sqrt{3}}{2} = \frac{L}{2\sqrt{3}}$.
The net force on $m_{2}$ is the force due to the mass $m_{1}$ at $A$:
$F = G \frac{m_{1} m_{2}}{h^{2}} = G \frac{m_{1} m_{2}}{\left(\frac{L}{2\sqrt{3}}\right)^{2}} = G \frac{m_{1} m_{2}}{\frac{L^{2}}{12}} = \frac{12 G m_{1} m_{2}}{L^{2}}$.

(B) The gravitational force is given by $F = \frac{G M m}{r^2}$.
Acceleration due to gravity is $g = \frac{F}{m} = \frac{G M}{r^2}$.
Therefore,the ratio $\frac{G}{g}$ can be expressed as:
$\frac{G}{g} = \frac{G}{\frac{G M}{r^2}} = \frac{r^2}{M}$.
The unit of $r$ is $m$ (meters) and the unit of $M$ is $kg$ (kilograms).
Thus,the $SI$ unit of $\frac{G}{g}$ is $\frac{m^2}{kg}$.

(B) The gravitational force between two masses is given by $F = \frac{G m_1 m_2}{r^2}$.
Given $m_A = m_B = 8 \,kg$, $m_G = 2 \,kg$, and $AG = BG = 1 \,m$.
The force exerted by mass at $A$ on mass at $G$ is $F_A = \frac{G \times 8 \times 2}{1^2} = 16G$.
The force exerted by mass at $B$ on mass at $G$ is $F_B = \frac{G \times 8 \times 2}{1^2} = 16G$.
The angle between $F_A$ and $F_B$ is $120^{\circ}$.
The resultant force $F_{AB}$ is given by $F_{AB} = \sqrt{F_A^2 + F_B^2 + 2 F_A F_B \cos 120^{\circ}}$.
Since $F_A = F_B = 16G$, $F_{AB} = \sqrt{(16G)^2 + (16G)^2 + 2(16G)(16G)(-0.5)} = \sqrt{3(16G)^2 - (16G)^2} = 16G$.
This resultant force $F_{AB}$ acts along the line $GC$ directed towards $C$.
To make the net force on the $2 \,kg$ body zero, a fourth body of mass $m_C = 4 \,kg$ must be placed at a distance $x$ from $G$ along the line $GC$ such that the gravitational force $F_C$ exerted by it balances $F_{AB}$.
$F_C = \frac{G \times 4 \times 2}{x^2} = 16G$.
$\frac{8G}{x^2} = 16G \Rightarrow x^2 = \frac{8}{16} = 0.5$.
$x = \frac{1}{\sqrt{2}} \,m$.
Thus, the fourth body should be placed at a point $P$ on the line $CG$ such that $PG = \frac{1}{\sqrt{2}} \,m$.

(B) Mass is a fundamental property of matter and it does not depend on external conditions such as gravity or location.
It represents the amount of matter contained in an object.
Since mass is an intrinsic property,it remains constant regardless of where the object is placed.
Therefore,if the mass of the body on the surface of the Earth is $M$,its mass on the surface of the moon will also be $M$.

(A) The relative strengths of fundamental forces are compared by considering their interaction between two protons at a distance of $10^{-15} \ m$.
The gravitational force $(F_G)$ is approximately $10^{-36} \ N$ to $10^{-39} \ N$ relative to the strong nuclear force $(F_S)$.
Specifically,the ratio of gravitational force to strong nuclear force is approximately $10^{-39}$.
Thus,the gravitational force is $10^{-39}$ times the strength of the strong nuclear force.

(A) The mass $M$ of a solid sphere of radius $R$ and density $\rho$ is given by $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$.
Since both spheres are made of the same material,their density $\rho$ is constant.
Thus,$M \propto R^3$.
The distance between the centers of the two spheres in contact is $d = R + R = 2R$.
The gravitational force $F$ between two spheres is given by Newton's Law of Gravitation: $F = \frac{G M_1 M_2}{d^2}$.
Substituting $M_1 = M_2 = M$ and $d = 2R$,we get $F = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2}$.
Since $M \propto R^3$,then $M^2 \propto (R^3)^2 = R^6$.
Substituting this into the force equation: $F \propto \frac{R^6}{R^2} = R^4$.
Therefore,$F \propto R^4$.

(D) The universal law of gravitation in scalar form is given by $F = G \frac{m_1 m_2}{r^2}$.
To express this in vector form,we multiply the magnitude by the unit vector $\hat{r}$ in the direction of the force: $\overrightarrow{F} = G \frac{m_1 m_2}{r^2} \hat{r}$.
Since the unit vector is defined as $\hat{r} = \frac{\overrightarrow{r}}{r}$,we substitute this into the equation:
$\overrightarrow{F} = G \frac{m_1 m_2}{r^2} \left( \frac{\overrightarrow{r}}{r} \right)$.
Therefore,the vector form is $\overrightarrow{F} = G \frac{m_1 m_2}{r^3} \overrightarrow{r}$.

(D) Given: Mass of each stone, $m_1 = m_2 = 2 \,kg$.
Separation distance, $r = 1 \,m$.
Universal gravitational constant, $G = 6.67 \times 10^{-11} \,N \cdot m^2/kg^2$.
According to Newton's law of gravitation, the force $F$ is given by $F = \frac{G m_1 m_2}{r^2}$.
Substituting the values: $F = \frac{6.67 \times 10^{-11} \times 2 \times 2}{1^2}$.
$F = 6.67 \times 10^{-11} \times 4$.
$F = 26.68 \times 10^{-11} \,N = 2.668 \times 10^{-10} \,N$.
Rounding to the nearest value, $F \approx 2.67 \times 10^{-10} \,N$.

(D) Let a unit mass $m$ be placed at a distance $x$ from the centre of the earth,where the net gravitational force is zero.
At this point,the gravitational pull from the earth must be equal in magnitude to the gravitational pull from the moon.
$\frac{G m M_e}{x^2} = \frac{G m M_m}{(D-x)^2}$ ... $(i)$
where $M_e$ is the mass of the earth and $M_m$ is the mass of the moon.
Given that $M_e = 81 M_m$.
Substituting this into equation $(i)$:
$\frac{G m (81 M_m)}{x^2} = \frac{G m M_m}{(D-x)^2}$
$\frac{81}{x^2} = \frac{1}{(D-x)^2}$
Taking the square root on both sides:
$\frac{9}{x} = \frac{1}{D-x}$
$9(D - x) = x$
$9D - 9x = x$
$9D = 10x$
$x = \frac{9D}{10}$
Thus,the gravitational force is zero at a distance of $\frac{9D}{10}$ from the centre of the earth.

(A) The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
Here,$m_1 = xM$ and $m_2 = (1-x)M$.
Substituting these,we get $F = \frac{G}{r^2} (xM)(1-x)M = \frac{GM^2}{r^2} (x - x^2)$.
For $F$ to be maximum,the derivative with respect to $x$ must be zero: $\frac{dF}{dx} = 0$.
$\frac{d}{dx} [\frac{GM^2}{r^2} (x - x^2)] = 0$.
Since $\frac{GM^2}{r^2}$ is constant,we have $\frac{d}{dx} (x - x^2) = 0$.
$1 - 2x = 0$.
Therefore,$x = 1/2$.

(C) Let the side length of the square be $a$. The net gravitational force on any one mass (say mass $4$) is the vector sum of the forces exerted by the other three masses $(1, 2, 3)$.
Let $\vec{F}_{14}$,$\vec{F}_{34}$,and $\vec{F}_{24}$ be the forces exerted by masses $1, 3,$ and $2$ on mass $4$ respectively.
The magnitudes are $F_{14} = \frac{Gm^2}{a^2}$,$F_{34} = \frac{Gm^2}{a^2}$,and $F_{24} = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$.
The resultant of $\vec{F}_{14}$ and $\vec{F}_{34}$ is $F_{13} = \sqrt{F_{14}^2 + F_{34}^2} = \sqrt{\left(\frac{Gm^2}{a^2}\right)^2 + \left(\frac{Gm^2}{a^2}\right)^2} = \sqrt{2} \frac{Gm^2}{a^2}$.
This resultant force $F_{13}$ acts along the diagonal,in the same direction as $\vec{F}_{24}$.
Thus,the net force is $F_{net} = F_{13} + F_{24} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left(\sqrt{2} + \frac{1}{2}\right) = \frac{Gm^2}{a^2} \left(\frac{2\sqrt{2} + 1}{2}\right)$.
Given that $F_{net} = \left(\frac{2\sqrt{2} + 1}{32}\right) \frac{Gm^2}{L^2}$,we equate the two expressions:
$\frac{Gm^2}{a^2} \left(\frac{2\sqrt{2} + 1}{2}\right) = \left(\frac{2\sqrt{2} + 1}{32}\right) \frac{Gm^2}{L^2}$.
$\frac{1}{2a^2} = \frac{1}{32L^2} \Rightarrow a^2 = 16L^2 \Rightarrow a = 4L$.

(C) The gravitational force $F$ between two masses $m_0$ and $M-m_0$ separated by a distance $D$ is given by:
$F = \frac{G m_0 (M - m_0)}{D^2}$
To find the condition for maximum force,we differentiate $F$ with respect to $m_0$ and set it to zero:
$\frac{dF}{dm_0} = \frac{d}{dm_0} \left[ \frac{G}{D^2} (M m_0 - m_0^2) \right] = 0$
$\frac{G}{D^2} (M - 2m_0) = 0$
Since $\frac{G}{D^2} \neq 0$,we have:
$M - 2m_0 = 0$
$M = 2m_0$
$\frac{m_0}{M} = \frac{1}{2} = 0.5$

(C) According to the Shell Theorem,the gravitational field $E$ inside a uniform hollow spherical shell is $0$. Therefore,no net gravitational force acts on a point mass placed inside it. Thus,Statement $(I)$ is false.
For any point outside a uniform hollow spherical shell,the shell behaves as if its entire mass were concentrated at its center. Thus,the gravitational force is calculated using Newton's Law of Gravitation as $F = G M m / r^2$,where $r$ is the distance from the center. Thus,Statement $(II)$ is correct.

(A) According to Newton's Law of Universal Gravitation,every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
This force of mutual attraction that exists between any two objects due to their masses is known as the Gravitational force.

(C) The gravitational force between two point masses is given by Newton's Law of Gravitation,$F = G \frac{m_1 m_2}{r^2}$.
$1$. It is a conservative force,meaning the work done by it is independent of the path.
$2$. It is always an attractive force between two masses.
$3$. It is a central force,as it acts along the line joining the centers of the two bodies.
$4$. It is a non-contact force (or action-at-a-distance force),meaning it does not require physical contact between the bodies.
Therefore,the statement that it is 'Not a central force' is incorrect.

(C) Among the four fundamental forces in nature (gravitational,electromagnetic,strong nuclear,and weak nuclear forces),the gravitational force is unique because it is always attractive in nature.
Electromagnetic forces can be either attractive or repulsive depending on the charges.
Strong nuclear forces are generally attractive at short ranges to hold the nucleus together.
Therefore,the statement that gravitational forces are always attractive is correct.