Gravitational Potential and Potential Energy of system Questions in English

(B) Gravitational potential $V$ at a point is defined as the work done per unit mass in bringing a test mass from infinity to that point.
Mathematically,$V = \frac{W}{m}$,where $W$ is work done and $m$ is mass.
The $S.I.$ unit of work $W$ is Joule $(J)$ and the $S.I.$ unit of mass $m$ is kilogram $(kg)$.
Therefore,the $S.I.$ unit of gravitational potential is $\frac{J}{kg} = J \cdot kg^{-1}$.

(B) The gravitational force is a conservative force.
By definition,the work done by or against a conservative force in moving a particle between two points depends only on the initial and final positions of the particle,not on the path taken.
Since all three paths $1, 2$ and $3$ start at point $A$ and end at point $B$,the work done along each path must be equal.
Therefore,$W_1 = W_2 = W_3$.

(C) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.
At height $h$,the potential energy is $U_f = -\frac{GMm}{R+h}$.
The increase in potential energy is $\Delta U = U_f - U_i = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = GMm \left( \frac{h}{R(R+h)} \right)$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting $GM$,we get $\Delta U = \frac{gR^2 mh}{R(R+h)} = \frac{mgh}{1 + h/R}$.
Given $h = R/5$,so $h/R = 1/5$.
Therefore,$\Delta U = \frac{mgh}{1 + 1/5} = \frac{mgh}{6/5} = \frac{5}{6}mgh$.

(B) The relationship between the gravitational field $I$ and the gravitational potential $V$ is given by the formula $I = -\frac{dV}{dr}$.
This implies that the gravitational field is the negative gradient of the gravitational potential.
However,it is a common misconception that $V = 0$ implies $I = 0$. For example,consider the gravitational potential at the center of a spherical shell or between two point masses; the potential can be zero at a point where the field is non-zero.
Therefore,the gravitational field is not necessarily zero just because the gravitational potential is zero.

(D) The gravitational potential $V(x)$ is defined as the negative integral of the gravitational field $E$ from infinity to the point $x$:
$V(x) = -\int_{\infty}^{x} E \, dx$
Given $E = K/x^3$,we have:
$V(x) = -\int_{\infty}^{x} \frac{K}{x^3} \, dx$
$V(x) = -K \int_{\infty}^{x} x^{-3} \, dx$
$V(x) = -K \left[ \frac{x^{-2}}{-2} \right]_{\infty}^{x}$
$V(x) = \frac{K}{2} \left[ \frac{1}{x^2} \right]_{\infty}^{x}$
$V(x) = \frac{K}{2} \left( \frac{1}{x^2} - 0 \right) = \frac{K}{2x^2}$

(A) The gravitational potential energy $U$ of a system of two masses $M$ and $m$ separated by a distance $r$ is given by the formula: $U = -\frac{GMm}{r}$.
Given:
$M = 6.00 \times 10^{24} \ kg$
$m = 7.40 \times 10^{22} \ kg$
$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$
$U = -7.79 \times 10^{28} \ J$
Rearranging the formula to solve for $r$:
$r = -\frac{GMm}{U}$
Substituting the values:
$r = \frac{6.67 \times 10^{-11} \times 6.00 \times 10^{24} \times 7.40 \times 10^{22}}{7.79 \times 10^{28}}$
$r = \frac{296.196 \times 10^{35}}{7.79 \times 10^{28}}$
$r \approx 38.02 \times 10^7 \ m = 3.80 \times 10^8 \ m$.

(B) The gravitational potential energy $U$ of a mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
At the surface of the Earth,$r = R$ and $m = 1 \, kg$,so the initial potential energy is $U_i = -\frac{GM(1)}{R} = -\frac{GM}{R}$.
At infinity,the distance $r = \infty$,so the final potential energy is $U_f = -\frac{GM(1)}{\infty} = 0$.
The work done $W$ is equal to the change in potential energy: $W = U_f - U_i$.
$W = 0 - (-\frac{GM}{R}) = \frac{GM}{R}$.

(D) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the earth's surface,$r = R_e$,so $U_1 = -\frac{GMm}{R_e} = -mgR_e$ (since $g = \frac{GM}{R_e^2}$).
At a height $h = R_e$ from the surface,the distance from the center is $r = R_e + h = R_e + R_e = 2R_e$.
The potential energy at this height is $U_2 = -\frac{GMm}{2R_e}$.
Substituting $GM = gR_e^2$,we get $U_2 = -\frac{(gR_e^2)m}{2R_e} = -\frac{1}{2}mgR_e$.

(B) The relationship between gravitational field $I$ and gravitational potential $V$ is given by the formula $I = -\frac{dV}{dr}$.
Given that the gravitational field $I = 0$ in the region.
Substituting this into the formula,we get $0 = -\frac{dV}{dr}$,which implies $\frac{dV}{dr} = 0$.
Since the derivative of the potential with respect to distance is zero,the gravitational potential $V$ must be constant throughout that region.

(A) The distance from each vertex of the square to its centre is $r = \frac{L}{\sqrt{2}}$.
Gravitational potential $V$ due to a single mass $M$ at a distance $r$ is given by $V = -\frac{GM}{r}$.
Substituting $r = \frac{L}{\sqrt{2}}$,the potential due to one particle is $V_1 = -\frac{GM}{L/\sqrt{2}} = -\frac{\sqrt{2}GM}{L}$.
Since there are four such particles,the total gravitational potential at the centre is $V_{total} = 4 \times V_1$.
$V_{total} = 4 \times \left( -\frac{\sqrt{2}GM}{L} \right) = -\frac{4\sqrt{2}GM}{L}$.
Since $4\sqrt{2} = \sqrt{16 \times 2} = \sqrt{32}$,the total potential is $V_{total} = -\sqrt{32} \frac{GM}{L}$.

(A) The potential energy $U$ of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
Here,$r = R_e + h$,where $R_e$ is the radius of the Earth and $h$ is the height of the satellite.
Given $h = 6.4 \times 10^6 \ m$ and $R_e \approx 6.4 \times 10^6 \ m$,we have $r = R_e + R_e = 2R_e$.
Using the relation $g = \frac{GM}{R_e^2}$,we can write $GM = gR_e^2$.
Substituting these into the potential energy formula:
$U = -\frac{(gR_e^2)m}{2R_e} = -\frac{1}{2}mgR_e = -0.5 \, mgR_e$.

(C) The gravitational potential $V$ of a spherical body of mass $M$ and radius $R$ is given by:
$1$. Inside the Earth $(r < R)$: $V_{in} = -\frac{GM}{2R^3} (3R^2 - r^2) = -\frac{GM}{2R} [3 - (r/R)^2]$. This is a parabolic variation where the potential is maximum (least negative) at the centre $(r=0)$ and decreases to $-\frac{3GM}{2R}$ at the surface.
$2$. At the surface $(r = R)$: $V_{surface} = -\frac{GM}{R}$.
$3$. Outside the Earth $(r > R)$: $V_{out} = -\frac{GM}{r}$. This follows an inverse relationship.
Comparing these with the given graphs, the potential starts from a negative value at the center, decreases parabolically to a more negative value at the surface, and then increases towards zero as $r$ approaches infinity. Option $C$ correctly represents this behavior.

(C) For a hollow sphere of mass $M$ and radius $R$:
$1$. Inside the sphere $(r < R)$, the gravitational potential is constant and equal to the potential at the surface: $V_{in} = -\frac{GM}{R}$.
$2$. At the surface $(r = R)$, the potential is $V_{surface} = -\frac{GM}{R}$.
$3$. Outside the sphere $(r > R)$, the potential varies inversely with distance: $V_{out} = -\frac{GM}{r}$.
Since the potential is negative, it remains constant at a negative value inside the sphere and approaches zero as $r$ increases outside the sphere. Graph $C$ correctly represents this behavior, showing a constant negative value for $r \leq R$ and a curve following $V \propto -1/r$ for $r > R$.

(D) system is considered bound if its total mechanical energy $(E = U + E_k)$ is negative.
In the given graph,$U$ represents the potential energy (which is negative) and $E_k$ represents the kinetic energy (which is positive).
At points $A, B,$ and $C$,the magnitude of the negative potential energy $|U|$ is greater than the kinetic energy $E_k$,which implies that the total energy $E = U + E_k < 0$.
Therefore,the system is bound at points $A, B,$ and $C$.

(C) The gravitational potential energy $U(r)$ of a point mass $m$ at a distance $r$ from the center of a spherical shell of mass $M$ and radius $R$ is given by $U(r) = m \cdot V(r)$,where $V(r)$ is the gravitational potential due to the shell.
For $r < R$,the gravitational potential inside a spherical shell is constant and equal to $V = -\frac{GM}{R}$. Thus,$U(r) = -\frac{GMm}{R}$,which is a constant value.
For $r \geq R$,the gravitational potential is $V = -\frac{GM}{r}$. Thus,$U(r) = -\frac{GMm}{r}$,which varies inversely with $r$.
The graph of $U(r)$ versus $r$ will be a horizontal line for $r \leq R$ and a curve following the inverse law for $r > R$. This matches the graph shown in option $C$.

(D) The relationship between gravitational potential $V$ and gravitational field intensity $E$ is given by the formula $V = - \int E \, dx$.
Given $E = K/x^3$.
Substituting the value of $E$ in the integral:
$V = - \int (K/x^3) \, dx$
$V = -K \int x^{-3} \, dx$
$V = -K [x^{-2} / -2]$
$V = K / (2x^2)$.
Thus,the gravitational potential is $K / (2x^2)$.

(C) The gravitational intensity $I$ at a distance $r$ from the center of the Earth is given by $I = \frac{GM}{r^2}$.
The gravitational potential $V$ at a distance $r$ is given by $V = -\frac{GM}{r}$.
By taking the magnitude,we can relate the two: $V = I \times r$.
Given: $I = 6 \, N/kg$ and $r = 8000 \, km = 8 \times 10^6 \, m$.
Substituting the values: $V = 6 \times (8 \times 10^6) = 48 \times 10^6 = 4.8 \times 10^7 \, J/kg$.

(B) The gravitational potential $V$ at the origin due to a system of point masses is given by the sum of potentials due to individual masses:
$V = - \left[ \frac{Gm}{r_1} + \frac{Gm}{r_2} + \frac{Gm}{r_3} + \dots \right]$
Substituting the given positions $r_1 = 1, r_2 = 2, r_3 = 4, r_4 = 8, \dots$:
$V = - Gm \left[ \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right]$
The term in the bracket is an infinite geometric series with first term $a = 1$ and common ratio $r = 1/2$. The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
$S = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$
Therefore,$V = - Gm(2) = - 2Gm$.

(B) Let the point where the gravitational field intensity is zero be at a distance $x$ from mass $m$. The gravitational field intensity $E$ is given by $E = \frac{Gm}{x^2} - \frac{GM}{(d-x)^2} = 0$.
This implies $\frac{m}{x^2} = \frac{M}{(d-x)^2}$,so $\frac{\sqrt{m}}{x} = \frac{\sqrt{M}}{d-x}$.
Solving for $x$,we get $x = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d$ and $d-x = \frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d$.
The gravitational potential $V$ at this point is the sum of potentials due to both masses: $V = - \frac{Gm}{x} - \frac{GM}{d-x}$.
Substituting the values of $x$ and $d-x$: $V = - \frac{Gm}{\frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d} - \frac{GM}{\frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d}$.
$V = - \frac{G}{d} [\sqrt{m}(\sqrt{m} + \sqrt{M}) + \sqrt{M}(\sqrt{m} + \sqrt{M})]$.
$V = - \frac{G}{d} (\sqrt{m} + \sqrt{M})^2$.

(A) The gravitational potential energy $U$ of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
Given height $h = 6.4 \times 10^6 \, m$. Since the radius of the Earth $R_e \approx 6.4 \times 10^6 \, m$,we have $h = R_e$.
The distance from the center of the Earth is $r = R_e + h = R_e + R_e = 2R_e$.
Substituting this into the formula: $U = -\frac{GMm}{2R_e}$.
Using the relation $GM = gR_e^2$,where $g$ is the acceleration due to gravity at the Earth's surface:
$U = -\frac{(gR_e^2)m}{2R_e} = -\frac{1}{2} mgR_e = -0.5 \, mgR_e$.

(C) The gravitational potential $V$ at a point due to a point mass $m$ at a distance $r$ is given by $V = -\frac{Gm}{r}$.
For a system of multiple masses,the total potential is the algebraic sum of the potentials due to each mass.
Here,each body has a mass $m = 2 \, kg$ and they are located at distances $r_1 = 1 \, m, r_2 = 2 \, m, r_3 = 4 \, m, r_4 = 8 \, m, \dots$ from the origin.
The total gravitational potential $V$ at the origin is:
$V = -\frac{G(2)}{1} - \frac{G(2)}{2} - \frac{G(2)}{4} - \frac{G(2)}{8} - \dots$
$V = -2G \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right)$
The term in the bracket is an infinite geometric series with the first term $a = 1$ and common ratio $r = \frac{1}{2}$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
$S = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
Therefore,$V = -2G(2) = -4G$.

(B) The gravitational potential energy $(U)$ of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$,where $G$ is the gravitational constant and $M$ is the mass of the Earth.
As the satellite moves to a further stable circular orbit,the orbital radius $r$ increases.
Since $U = -\frac{GMm}{r}$,as $r$ increases,the magnitude of the negative value decreases,which means the value of $U$ increases (becomes less negative).
Therefore,the gravitational potential energy increases.

(D) The gravitational potential $V$ at a distance $r$ from the center of the Earth is given by $V = -\frac{GM}{r}$.
The gravitational field intensity $I$ at a distance $r$ from the center of the Earth is given by $I = \frac{GM}{r^2}$.
For the gravitational potential $V$ to be zero,we require $r \to \infty$.
For the gravitational field $I$ to be zero,we require $r \to \infty$.
Therefore,both the gravitational potential and the gravitational field due to the Earth are zero only at infinity.

(C) By the law of conservation of energy,the initial potential energy at infinity is $0$ and the initial kinetic energy is $0$. At a separation distance $d$,the potential energy is $-G m_1 m_2 / d$. The total energy is conserved:
$0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 - \frac{G m_1 m_2}{d}$
From the conservation of linear momentum,since the system is initially at rest,the center of mass remains at rest:
$m_1 v_1 = m_2 v_2 \implies v_2 = \frac{m_1 v_1}{m_2}$
Substituting $v_2$ into the energy equation:
$\frac{G m_1 m_2}{d} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 (\frac{m_1 v_1}{m_2})^2 = \frac{1}{2} m_1 v_1^2 (1 + \frac{m_1}{m_2}) = \frac{1}{2} m_1 v_1^2 (\frac{m_1 + m_2}{m_2})$
Solving for $v_1$: $v_1 = \sqrt{\frac{2 G m_2^2}{d(m_1 + m_2)}}$. Similarly,$v_2 = \sqrt{\frac{2 G m_1^2}{d(m_1 + m_2)}}$.
The relative velocity of approach is $v_{rel} = v_1 + v_2 = \sqrt{\frac{2 G}{d(m_1 + m_2)}} (m_2 + m_1) = \sqrt{\frac{2 G (m_1 + m_2)}{d}}$.

(B) The gravitational potential $V$ at the centre of a hollow spherical shell of mass $M$ and radius $r$ is given by $V = -\frac{GM}{r}$.
When the shell is compressed to half its radius,the new radius becomes $r' = \frac{r}{2}$.
The new potential at the centre becomes $V' = -\frac{GM}{r'} = -\frac{GM}{r/2} = -\frac{2GM}{r} = 2V$.
Since $V$ is negative,$2V$ is more negative than $V$ (e.g.,if $V = -10 \ J/kg$,then $V' = -20 \ J/kg$).
Therefore,the gravitational potential decreases.

(C) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.
For figure $(i)$,the distances are $r_{m,2m} = a$,$r_{m,3m} = a$,and $r_{2m,3m} = \sqrt{a^2 + a^2} = a\sqrt{2}$.
$U_1 = -\left( \frac{G(m)(2m)}{a} + \frac{G(m)(3m)}{a} + \frac{G(2m)(3m)}{a\sqrt{2}} \right) = -\frac{Gm^2}{a} \left( 2 + 3 + \frac{6}{\sqrt{2}} \right) = -\frac{Gm^2}{a} (5 + 3\sqrt{2})$.
For figure $(ii)$,the distances are $r_{m,2m} = a$,$r_{m,3m} = a$,and $r_{2m,3m} = a$.
$U_2 = -\left( \frac{G(m)(2m)}{a} + \frac{G(m)(3m)}{a} + \frac{G(2m)(3m)}{a} \right) = -\frac{Gm^2}{a} (2 + 3 + 6) = -\frac{11Gm^2}{a}$.
The work done by an external agent is $W = U_2 - U_1$.
$W = -\frac{11Gm^2}{a} - \left( -\frac{Gm^2}{a} (5 + 3\sqrt{2}) \right) = \frac{Gm^2}{a} (-11 + 5 + 3\sqrt{2}) = \frac{Gm^2}{a} (-6 + 3\sqrt{2})$.
$W = -\frac{6Gm^2}{a} \left( 1 - \frac{3\sqrt{2}}{6} \right) = -\frac{6Gm^2}{a} \left( 1 - \frac{1}{\sqrt{2}} \right)$.

(D) The gravitational potential $V$ due to a mass element $dm$ at a distance $r$ is given by $V = -\int \frac{G dm}{r}.$
For a hemispherical bowl of radius $R$ and mass $M,$ every point on the surface is at a distance $R$ from the centre. Thus,$V = -\int \frac{G dm}{R} = -\frac{G}{R} \int dm = -\frac{GM}{R}.$
For a thin uniform wire of mass $M$ bent into a semicircle of radius $R,$ every point on the wire is at a distance $R$ from the centre of curvature. Thus,$V = -\int \frac{G dm}{R} = -\frac{GM}{R}.$
Since the distance $R$ is constant for all mass elements in both cases,the potential remains $-GM/R.$
This holds true regardless of the distribution of mass along the wire,as long as every element is at distance $R.$ Therefore,both $(A)$ and $(C)$ are correct.

(A) The gravitational potential energy of a satellite in an orbit of radius $r$ is given by $U = -\frac{GMm}{r}$. As the radius $r$ increases,the value of $U$ becomes less negative,meaning it increases.
The orbital velocity is $v = \sqrt{\frac{GM}{r}}$. As $r$ increases,$v$ decreases.
The angular velocity is $\omega = \frac{v}{r} = \sqrt{\frac{GM}{r^3}}$. As $r$ increases,$\omega$ decreases.
The centripetal acceleration is $a_c = \frac{v^2}{r} = \frac{GM}{r^2}$. As $r$ increases,$a_c$ decreases.
Therefore,only the gravitational potential energy increases.

(C) Let the gravitational field at point $P$,at a distance $x$ from mass $m$,be zero.
Equating the gravitational field strengths due to both masses:
$\frac{Gm}{x^2} = \frac{G(4m)}{(r-x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{r-x}$
$r - x = 2x \implies 3x = r \implies x = \frac{r}{3}$
Now,calculate the gravitational potential $V$ at point $P$:
$V = -\frac{Gm}{x} - \frac{G(4m)}{r-x}$
Substituting $x = \frac{r}{3}$ and $r-x = \frac{2r}{3}$:
$V = -\frac{Gm}{r/3} - \frac{4Gm}{2r/3} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}$

(A) Let the center of the original sphere be $O$ and the center of the cavity be $P$. The distance $OP = R/2$.
The gravitational potential at any point inside a solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center is given by $V = \frac{-GM}{2R^3}(3R^2 - r^2)$.
$1$. Potential at point $P$ due to the original solid sphere:
Here,$r = OP = R/2$.
$V_{sphere} = \frac{-GM}{2R^3} \left[ 3R^2 - (R/2)^2 \right] = \frac{-GM}{2R^3} \left( 3R^2 - \frac{R^2}{4} \right) = \frac{-GM}{2R^3} \left( \frac{11R^2}{4} \right) = \frac{-11GM}{8R}$.
$2$. Potential at point $P$ due to the removed spherical portion (cavity):
The mass of the removed portion $M'$ is proportional to the volume: $M' = M \times \frac{(4/3)\pi(R/2)^3}{(4/3)\pi R^3} = M/8$.
The potential at the center of a solid sphere of mass $M'$ and radius $R' = R/2$ is $V_{cavity} = \frac{-3GM'}{2R'} = \frac{-3G(M/8)}{2(R/2)} = \frac{-3GM}{8R}$.
$3$. Potential at the center of the cavity $(P)$:
$V_{total} = V_{sphere} - V_{cavity} = \frac{-11GM}{8R} - \left( \frac{-3GM}{8R} \right) = \frac{-11GM + 3GM}{8R} = \frac{-8GM}{8R} = \frac{-GM}{R}$.

(A) The gravitational potential energy at the center of the Earth is given by $U_i = -\frac{3GMm}{2R}$.
The gravitational potential energy at the surface of the Earth is given by $U_f = -\frac{GMm}{R}$.
The energy required to move the mass is the change in potential energy: $\Delta U = U_f - U_i = -\frac{GMm}{R} - (-\frac{3GMm}{2R}) = \frac{GMm}{2R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the equation:
$\Delta U = \frac{(gR^2)m}{2R} = \frac{mgR}{2}$.
Given $m = 4 \, kg$,$g = 10 \, m/s^2$,and $R = 6400 \, km = 6.4 \times 10^6 \, m$:
$\Delta U = \frac{4 \times 10 \times 6.4 \times 10^6}{2} = 2 \times 6.4 \times 10^7 = 1.28 \times 10^8 \, J$.

(D) The gravitational potential energy $U$ of a uniform spherical star of mass $M$ and radius $R$ is given by $U = -\frac{3GM^2}{5R}$.
The volume $V$ of the star is $V = \frac{4}{3}\pi R^3$,which implies $R = (\frac{3V}{4\pi})^{1/3}$.
Substituting $R$ into the expression for $U$,we get $U = -\frac{3GM^2}{5} (\frac{4\pi}{3V})^{1/3} = -\frac{3GM^2}{5} (\frac{4\pi}{3})^{1/3} V^{-1/3}$.
The gravitational pressure $P$ is related to the change in potential energy by the relation $dW = P dV = -dU$,which gives $P = -\frac{dU}{dV}$.
Differentiating $U$ with respect to $V$: $P = -[ -\frac{3GM^2}{5} (\frac{4\pi}{3})^{1/3} \cdot (-\frac{1}{3}) V^{-4/3} ]$.
Simplifying this,we find $P = -\frac{GM^2}{5} (\frac{4\pi}{3})^{1/3} V^{-4/3}$.
Since $G, M,$ and the constants are fixed,$P \propto V^{-4/3}$.

(B) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{Gm_i m_j}{r_{ij}}$.
For case $(A)$,we have three pairs of masses with distances $d, d,$ and $\sqrt{d^2+d^2} = d\sqrt{2}$.
$U_A = -\frac{Gm^2}{d} - \frac{Gm^2}{d} - \frac{Gm^2}{d\sqrt{2}} = -\frac{2Gm^2}{d} - \frac{Gm^2}{d\sqrt{2}}$.
For case $(B)$,we have three pairs of masses with distances $d, d,$ and $2d$.
$U_B = -\frac{Gm^2}{d} - \frac{Gm^2}{d} - \frac{Gm^2}{2d} = -\frac{2Gm^2}{d} - \frac{Gm^2}{2d}$.
Comparing the two,since $\frac{1}{\sqrt{2}} \approx 0.707$ and $\frac{1}{2} = 0.5$,we have $\frac{1}{\sqrt{2}} > \frac{1}{2}$.
Therefore,$-\frac{Gm^2}{d\sqrt{2}} < -\frac{Gm^2}{2d}$.
This implies $U_A < U_B$.

(C) The total mechanical energy of the particle is conserved.
At infinity,the potential energy is $U_{\infty} = 0$ and the kinetic energy is $K_{\infty} = 0$.
At the centre of the sphere,the potential energy $U_{c}$ is given by the formula for the potential inside a uniform solid sphere: $U_{c} = m \times V_{c} = m \times \left( -\frac{3GM}{2R} \right) = -\frac{3GMm}{2R}$.
Applying the law of conservation of energy: $K_{\infty} + U_{\infty} = K_{c} + U_{c}$.
$0 + 0 = \frac{1}{2}mV^{2} - \frac{3GMm}{2R}$.
$\frac{1}{2}mV^{2} = \frac{3GMm}{2R}$.
$V^{2} = \frac{3GM}{R}$.
$V = \sqrt{\frac{3GM}{R}}$.

(D) The gravitational potential $V$ at a distance $r$ from a point mass $m$ is given by $V = -\frac{Gm}{r}$.
For a continuous distribution of mass,the potential at the center of a circular arc is the same for every point on the arc because every point is at the same distance $r$ from the center.
Given that the length of the rod is $L$ and it is bent into a semicircle,the radius $r$ of the semicircle is related to the length by the formula for the circumference of a semicircle: $\pi r = L$.
Therefore,the radius is $r = \frac{L}{\pi}$.
Since the total mass $M$ is distributed uniformly along the arc,the gravitational potential at the center is $V = -\frac{GM}{r}$.
Substituting the value of $r$,we get $V = -\frac{GM}{(L/\pi)}$.
Thus,the gravitational potential at the center is $V = -\frac{\pi GM}{L}$.

(B) The relationship between gravitational potential $V$ and gravitational field $E$ is given by $dV = -\vec{E} \cdot d\vec{r}$.
Given $E = -\frac{K}{r}$,we have $dV = -(-\frac{K}{r}) dr = \frac{K}{r} dr$.
Integrating both sides from the reference point $(r_0, V_0)$ to an arbitrary point $(r, V)$:
$\int_{V_0}^{V} dV = \int_{r_0}^{r} \frac{K}{r} dr$.
$V - V_0 = K [\ln(r)]_{r_0}^{r}$.
$V - V_0 = K (\ln(r) - \ln(r_0))$.
$V - V_0 = K \ln \left( \frac{r}{r_0} \right)$.
Therefore,$V = V_0 + K \ln \left( \frac{r}{r_0} \right)$.

(C) We know that the gravitational field intensity $E$ is related to the gravitational potential $V$ by the relation $E = -dV/dr$.
Given $E = -k/r$,we have $-dV/dr = -k/r$,which simplifies to $dV/dr = k/r$.
Integrating both sides with respect to $r$ from the reference point $r_0$ to $r$:
$\int_{V_0}^{V} dV = \int_{r_0}^{r} (k/r) dr$
$V - V_0 = k [\ln(r)]_{r_0}^{r}$
$V - V_0 = k (\ln(r) - \ln(r_0))$
$V - V_0 = k \ln(r/r_0)$
Therefore,$V = V_0 + k \ln(r/r_0)$.

(B) The total gravitational potential at the surface of the shell is the sum of the potential due to the spherical shell and the potential due to the particle at the centre.
$1$. The potential due to a uniform spherical shell of mass $2M$ and radius $R$ at its surface is $V_{\text{shell}} = -\frac{G(2M)}{R}$.
$2$. The potential due to a particle of mass $M$ at a distance $R$ from it (which is the distance from the centre to the surface) is $V_M = -\frac{GM}{R}$.
$3$. The total potential $V_{\text{surface}}$ is given by:
$V_{\text{surface}} = V_{\text{shell}} + V_M$
$V_{\text{surface}} = -\frac{2GM}{R} - \frac{GM}{R}$
$V_{\text{surface}} = -\frac{3GM}{R}$

(A) The gravitational potential $V$ inside the Earth at a distance $r$ from the center is given by $V = -\frac{GM}{2R^3}(3R^2 - r^2)$.
At the surface $(r = R)$,the potential is $V_s = -\frac{GM}{R}$.
At a distance $r = \frac{R}{2}$ from the center,the potential is $V_p = -\frac{GM}{2R^3}(3R^2 - (\frac{R}{2})^2) = -\frac{GM}{2R^3}(3R^2 - \frac{R^2}{4}) = -\frac{GM}{2R^3}(\frac{11R^2}{4}) = -\frac{11GM}{8R}$.
Using the law of conservation of energy between the surface and the point at $\frac{R}{2}$:
$K_i + U_i = K_f + U_f$
$0 + m V_s = \frac{1}{2}mv^2 + m V_p$
$m(-\frac{GM}{R}) = \frac{1}{2}mv^2 + m(-\frac{11GM}{8R})$
$-\frac{GM}{R} + \frac{11GM}{8R} = \frac{1}{2}v^2$
$\frac{3GM}{8R} = \frac{1}{2}v^2$
$v^2 = \frac{3GM}{4R}$
$v = \sqrt{\frac{3GM}{4R}}$

(C) For a solid sphere of radius $R$ and mass $M$ with uniform density,the gravitational potential $V(r)$ is given by:
Inside the sphere $(r \le R)$: $V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)$. This is a parabolic variation where the potential is maximum (least negative) at the surface $(r=R)$ and minimum (most negative) at the center $(r=0)$.
Outside the sphere $(r > R)$: $V(r) = -\frac{GM}{r}$. This shows an inverse variation with distance.
Graph $C$ correctly shows the potential starting from a negative value at the center,increasing (becoming less negative) as $r$ increases towards $R$,and then continuing to increase towards zero as $r$ increases beyond $R$.

(D) The gravitational field is given by $\vec{g} = (5\hat{i} + 12\hat{j})\,N/kg$.
Since the field is uniform,the change in gravitational potential $\Delta V$ is given by $\Delta V = -\int \vec{g} \cdot d\vec{r}$.
For a displacement from $(0, 0)$ to $(7, -3)$,the change in potential is $\Delta V = -\vec{g} \cdot \Delta\vec{r}$.
$\Delta\vec{r} = (7 - 0)\hat{i} + (-3 - 0)\hat{j} = 7\hat{i} - 3\hat{j}$.
$\Delta V = -[(5\hat{i} + 12\hat{j}) \cdot (7\hat{i} - 3\hat{j})] = -[5(7) + 12(-3)] = -[35 - 36] = -(-1) = 1\,J/kg$.
The change in gravitational potential energy $\Delta U = m \Delta V$.
Given $m = 1\,kg$,$\Delta U = 1\,kg \times 1\,J/kg = 1\,J$.

(C) Let $M$ be the mass of the particle.
Using the principle of conservation of energy,the total energy at the initial position (at distance $r$ from the center) equals the total energy at the center $C$.
The gravitational potential $V_p$ at a distance $x$ on the axis of a ring of mass $m$ and radius $r$ is given by $V_p = -\frac{Gm}{\sqrt{r^2 + x^2}}$.
Initial potential energy $U_i = M \times V_p(r) = -\frac{GMm}{\sqrt{r^2 + r^2}} = -\frac{GMm}{r\sqrt{2}}$.
Final potential energy at center $U_f = M \times V_p(0) = -\frac{GMm}{r}$.
By conservation of energy: $U_i + K_i = U_f + K_f$.
$-\frac{GMm}{r\sqrt{2}} + 0 = -\frac{GMm}{r} + \frac{1}{2}MV^2$.
$\frac{1}{2}MV^2 = \frac{GMm}{r} - \frac{GMm}{r\sqrt{2}} = \frac{GMm}{r} (1 - \frac{1}{\sqrt{2}})$.
$V^2 = \frac{2Gm}{r} (1 - \frac{1}{\sqrt{2}})$.
$V = \sqrt{\frac{2Gm}{r} (1 - \frac{1}{\sqrt{2}})}$.

(C) The gravitational potential at any point on the surface of the spherical shell is the sum of the potential due to the particle at the centre and the potential due to the shell itself.
$1$. The potential due to the particle of mass $m$ at a distance $R$ (on the surface) is $V_1 = -\frac{Gm}{R}$.
$2$. The potential due to a uniform spherical shell of mass $3m$ and radius $R$ at any point on its surface is $V_2 = -\frac{G(3m)}{R}$.
$3$. The total gravitational potential $V$ on the surface is $V = V_1 + V_2$.
$V = -\frac{Gm}{R} + \left(-\frac{3Gm}{R}\right) = -\frac{4Gm}{R}$.

(B) The gravitational potential energy of a system of particles is the sum of the potential energies of all possible pairs.
For a square of side $l$,there are $4$ sides of length $l$ and $2$ diagonals of length $\sqrt{2}l$.
Total potential energy $U = 4 \times (-\frac{Gm^2}{l}) + 2 \times (-\frac{Gm^2}{\sqrt{2}l})$.
$U = -\frac{4Gm^2}{l} - \frac{2Gm^2}{\sqrt{2}l}$.
Factoring out $-\frac{2Gm^2}{l}$,we get $U = -\frac{2Gm^2}{l} (2 + \frac{1}{\sqrt{2}})$.
Thus,the correct option is $B$.

(A) The gravitational potential difference $\Delta V$ for a height $h_1 = 20\,m$ is given as $2\,J/kg$.
Since the gravitational field $g$ is uniform,the potential difference is given by $\Delta V = g \cdot h$.
Therefore,$g = \frac{\Delta V}{h_1} = \frac{2\,J/kg}{20\,m} = 0.1\,N/kg$ (or $m/s^2$).
The work done $W$ in moving a body of mass $m = 5\,kg$ to a height $h_2 = 4\,m$ is given by $W = m \cdot g \cdot h_2$.
Substituting the values: $W = 5\,kg \times 0.1\,N/kg \times 4\,m = 2\,J$.

(D) The total gravitational potential $V$ at a distance $r = \frac{3}{2}R$ is the sum of the potential due to the solid sphere $(V_1)$ and the potential due to the spherical shell $(V_2)$.
$1$. For the solid sphere of mass $M$ and radius $R$,the point $r = \frac{3}{2}R$ is outside the sphere. The potential is given by $V_1 = -\frac{GM}{r} = -\frac{GM}{(3/2)R} = -\frac{2GM}{3R}$.
$2$. For the spherical shell of mass $M$ and radius $2R$,the point $r = \frac{3}{2}R$ is inside the shell. The potential inside a shell is constant and equal to the potential at its surface,given by $V_2 = -\frac{GM}{R_{shell}} = -\frac{GM}{2R}$.
$3$. The net gravitational potential is $V = V_1 + V_2 = -\frac{2GM}{3R} - \frac{GM}{2R} = -\frac{4GM + 3GM}{6R} = -\frac{7GM}{6R}$.

(B) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth (mass $M$) is given by $U = -\frac{GMm}{r}$.
Here,the body is at a height $h$ above the surface of the Earth,so its distance from the center is $r = R + h$.
Therefore,the potential energy is $U = -\frac{GMm}{R+h}$.
Note: If $h$ is very small compared to $R$,we can approximate this using the Taylor expansion: $U = -\frac{GMm}{R(1 + h/R)} \approx -\frac{GMm}{R}(1 - h/R) = -\frac{GMm}{R} + \frac{GMmh}{R^2}$.
Since $g = \frac{GM}{R^2}$,this becomes $U \approx -\frac{GMm}{R} + mgh$.

(B) The gravitational potential energy of a system of three particles of mass $M$ at the corners of an equilateral triangle of side $a$ is given by the sum of potential energies of all pairs: $U_1 = -\frac{GM^2}{a} - \frac{GM^2}{a} - \frac{GM^2}{a} = -\frac{3GM^2}{a}$.
When the side length is changed to $2a$,the new potential energy is $U_2 = -\frac{3GM^2}{2a}$.
The work done by an external agent is equal to the change in potential energy: $W = U_2 - U_1$.
$W = -\frac{3GM^2}{2a} - (-\frac{3GM^2}{a}) = -\frac{3GM^2}{2a} + \frac{3GM^2}{a} = \frac{3GM^2}{2a}$.

(C) For a point at distance $x$ from the centre,where $r < x < R$:
$1$. The point lies outside the inner shell (radius $r$,mass $m$). The potential due to the inner shell is $V_1 = -\frac{Gm}{x}$.
$2$. The point lies inside the outer shell (radius $R$,mass $M$). The potential inside a spherical shell is constant and equal to the potential at its surface,so $V_2 = -\frac{GM}{R}$.
$3$. The total gravitational potential $V$ is the sum of the potentials due to both shells: $V = V_1 + V_2 = -\frac{Gm}{x} - \frac{GM}{R} = -G \left[ \frac{M}{R} + \frac{m}{x} \right]$.