Explanation:

(A - Q) \(P(r)\) shows a constant value until \(r=R\) and then drops off. This represents the variation of gravitational potential \(P\) for a uniform spherical shell.
\((\mathrm{B}-\mathrm{R}) P(r)\) shows a linear increase with \(r\) until \(r=R\) and then drops to zero. This represents the variation of gravitational field intensity for a uniform solid sphere
(C - S) \(P(r)\) shows a spike at \(r=R\). This represents the variation of gravitational field intensity \(P\) for a uniform spherical shell. The gravitational field is zero inside a spherical shell and has a spike at the surface due to the mass distribution.
(D - P) \(P(r)\) shows an increase until \(r=R\) and then gravitational potential decreases. Correct option is (3).

Explanation:

Gravitation potential at a point is given by
\(V=\dfrac{-G M}{r}\) \(\quad \quad (1)\)
where \(r\) is the distance between the point and centre of mass \(M\) i.e., centre of earth
The acceleration due to gravity, \(g=\dfrac{G M}{r^{2}}\) \(\quad \quad (2)\)
Dividing eqn. (1) and eqn. (2), we have
\(\dfrac{V}{g}=\left(\dfrac{-G M}{r}\right)\left(\dfrac{r^{2}}{G M}\right)\)
\(\Rightarrow r=\dfrac{-V}{g}=\dfrac{-\left(-5.12 \times 10^{7}\right)}{6.4}\)
\( = 8 \times {10^6}\;m = 8000\;km\)
\(\therefore\) Height of the point from earth's surface,
\(h=r-R\)
\(\therefore h = 8000 - 6400 = 1600\;km\)
So, correct option is (1).

Explanation:

As all the points on the periphery of either ring are at same distance from point \(P\) the potential at point \(P\) due to whole ring can be calculated as \(V=-(G M)\left(\sqrt{R^{2}+x^{2}}\right)\) where \(x\) is axial distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform are nonuniform.
\(\begin{aligned}& V \text { at } P \text { is } V=-\dfrac{G M}{\sqrt{2 R}}-\dfrac{G 2 M}{\sqrt{5 R}} \\& =-\dfrac{G M}{R}\left[\dfrac{1}{\sqrt{2}}+\dfrac{2}{\sqrt{5}}\right] .\end{aligned}\)

Explanation:

\(V=\dfrac{-G m}{R}-\dfrac{G m}{2 R}=\dfrac{-3 G m}{2 R}\)

Explanation:

Explanation:

(A - Q) \(P(r)\) shows a constant value until \(r=R\) and then drops off. This represents the variation of gravitational potential \(P\) for a uniform spherical shell.
\((\mathrm{B}-\mathrm{R}) P(r)\) shows a linear increase with \(r\) until \(r=R\) and then drops to zero. This represents the variation of gravitational field intensity for a uniform solid sphere
(C - S) \(P(r)\) shows a spike at \(r=R\). This represents the variation of gravitational field intensity \(P\) for a uniform spherical shell. The gravitational field is zero inside a spherical shell and has a spike at the surface due to the mass distribution.
(D - P) \(P(r)\) shows an increase until \(r=R\) and then gravitational potential decreases. Correct option is (3).

Explanation:

Gravitation potential at a point is given by
\(V=\dfrac{-G M}{r}\) \(\quad \quad (1)\)
where \(r\) is the distance between the point and centre of mass \(M\) i.e., centre of earth
The acceleration due to gravity, \(g=\dfrac{G M}{r^{2}}\) \(\quad \quad (2)\)
Dividing eqn. (1) and eqn. (2), we have
\(\dfrac{V}{g}=\left(\dfrac{-G M}{r}\right)\left(\dfrac{r^{2}}{G M}\right)\)
\(\Rightarrow r=\dfrac{-V}{g}=\dfrac{-\left(-5.12 \times 10^{7}\right)}{6.4}\)
\( = 8 \times {10^6}\;m = 8000\;km\)
\(\therefore\) Height of the point from earth's surface,
\(h=r-R\)
\(\therefore h = 8000 - 6400 = 1600\;km\)
So, correct option is (1).

Explanation:

As all the points on the periphery of either ring are at same distance from point \(P\) the potential at point \(P\) due to whole ring can be calculated as \(V=-(G M)\left(\sqrt{R^{2}+x^{2}}\right)\) where \(x\) is axial distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform are nonuniform.
\(\begin{aligned}& V \text { at } P \text { is } V=-\dfrac{G M}{\sqrt{2 R}}-\dfrac{G 2 M}{\sqrt{5 R}} \\& =-\dfrac{G M}{R}\left[\dfrac{1}{\sqrt{2}}+\dfrac{2}{\sqrt{5}}\right] .\end{aligned}\)

Explanation:

\(V=\dfrac{-G m}{R}-\dfrac{G m}{2 R}=\dfrac{-3 G m}{2 R}\)

Explanation:

Explanation:

(A - Q) \(P(r)\) shows a constant value until \(r=R\) and then drops off. This represents the variation of gravitational potential \(P\) for a uniform spherical shell.
\((\mathrm{B}-\mathrm{R}) P(r)\) shows a linear increase with \(r\) until \(r=R\) and then drops to zero. This represents the variation of gravitational field intensity for a uniform solid sphere
(C - S) \(P(r)\) shows a spike at \(r=R\). This represents the variation of gravitational field intensity \(P\) for a uniform spherical shell. The gravitational field is zero inside a spherical shell and has a spike at the surface due to the mass distribution.
(D - P) \(P(r)\) shows an increase until \(r=R\) and then gravitational potential decreases. Correct option is (3).

Explanation:

Gravitation potential at a point is given by
\(V=\dfrac{-G M}{r}\) \(\quad \quad (1)\)
where \(r\) is the distance between the point and centre of mass \(M\) i.e., centre of earth
The acceleration due to gravity, \(g=\dfrac{G M}{r^{2}}\) \(\quad \quad (2)\)
Dividing eqn. (1) and eqn. (2), we have
\(\dfrac{V}{g}=\left(\dfrac{-G M}{r}\right)\left(\dfrac{r^{2}}{G M}\right)\)
\(\Rightarrow r=\dfrac{-V}{g}=\dfrac{-\left(-5.12 \times 10^{7}\right)}{6.4}\)
\( = 8 \times {10^6}\;m = 8000\;km\)
\(\therefore\) Height of the point from earth's surface,
\(h=r-R\)
\(\therefore h = 8000 - 6400 = 1600\;km\)
So, correct option is (1).

Explanation:

As all the points on the periphery of either ring are at same distance from point \(P\) the potential at point \(P\) due to whole ring can be calculated as \(V=-(G M)\left(\sqrt{R^{2}+x^{2}}\right)\) where \(x\) is axial distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform are nonuniform.
\(\begin{aligned}& V \text { at } P \text { is } V=-\dfrac{G M}{\sqrt{2 R}}-\dfrac{G 2 M}{\sqrt{5 R}} \\& =-\dfrac{G M}{R}\left[\dfrac{1}{\sqrt{2}}+\dfrac{2}{\sqrt{5}}\right] .\end{aligned}\)

Explanation:

\(V=\dfrac{-G m}{R}-\dfrac{G m}{2 R}=\dfrac{-3 G m}{2 R}\)

Explanation:

Explanation:

(A - Q) \(P(r)\) shows a constant value until \(r=R\) and then drops off. This represents the variation of gravitational potential \(P\) for a uniform spherical shell.
\((\mathrm{B}-\mathrm{R}) P(r)\) shows a linear increase with \(r\) until \(r=R\) and then drops to zero. This represents the variation of gravitational field intensity for a uniform solid sphere
(C - S) \(P(r)\) shows a spike at \(r=R\). This represents the variation of gravitational field intensity \(P\) for a uniform spherical shell. The gravitational field is zero inside a spherical shell and has a spike at the surface due to the mass distribution.
(D - P) \(P(r)\) shows an increase until \(r=R\) and then gravitational potential decreases. Correct option is (3).

Explanation:

Gravitation potential at a point is given by
\(V=\dfrac{-G M}{r}\) \(\quad \quad (1)\)
where \(r\) is the distance between the point and centre of mass \(M\) i.e., centre of earth
The acceleration due to gravity, \(g=\dfrac{G M}{r^{2}}\) \(\quad \quad (2)\)
Dividing eqn. (1) and eqn. (2), we have
\(\dfrac{V}{g}=\left(\dfrac{-G M}{r}\right)\left(\dfrac{r^{2}}{G M}\right)\)
\(\Rightarrow r=\dfrac{-V}{g}=\dfrac{-\left(-5.12 \times 10^{7}\right)}{6.4}\)
\( = 8 \times {10^6}\;m = 8000\;km\)
\(\therefore\) Height of the point from earth's surface,
\(h=r-R\)
\(\therefore h = 8000 - 6400 = 1600\;km\)
So, correct option is (1).

Explanation:

As all the points on the periphery of either ring are at same distance from point \(P\) the potential at point \(P\) due to whole ring can be calculated as \(V=-(G M)\left(\sqrt{R^{2}+x^{2}}\right)\) where \(x\) is axial distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform are nonuniform.
\(\begin{aligned}& V \text { at } P \text { is } V=-\dfrac{G M}{\sqrt{2 R}}-\dfrac{G 2 M}{\sqrt{5 R}} \\& =-\dfrac{G M}{R}\left[\dfrac{1}{\sqrt{2}}+\dfrac{2}{\sqrt{5}}\right] .\end{aligned}\)

Explanation:

\(V=\dfrac{-G m}{R}-\dfrac{G m}{2 R}=\dfrac{-3 G m}{2 R}\)

Explanation:

Explanation:

(A - Q) \(P(r)\) shows a constant value until \(r=R\) and then drops off. This represents the variation of gravitational potential \(P\) for a uniform spherical shell.
\((\mathrm{B}-\mathrm{R}) P(r)\) shows a linear increase with \(r\) until \(r=R\) and then drops to zero. This represents the variation of gravitational field intensity for a uniform solid sphere
(C - S) \(P(r)\) shows a spike at \(r=R\). This represents the variation of gravitational field intensity \(P\) for a uniform spherical shell. The gravitational field is zero inside a spherical shell and has a spike at the surface due to the mass distribution.
(D - P) \(P(r)\) shows an increase until \(r=R\) and then gravitational potential decreases. Correct option is (3).

Explanation:

Gravitation potential at a point is given by
\(V=\dfrac{-G M}{r}\) \(\quad \quad (1)\)
where \(r\) is the distance between the point and centre of mass \(M\) i.e., centre of earth
The acceleration due to gravity, \(g=\dfrac{G M}{r^{2}}\) \(\quad \quad (2)\)
Dividing eqn. (1) and eqn. (2), we have
\(\dfrac{V}{g}=\left(\dfrac{-G M}{r}\right)\left(\dfrac{r^{2}}{G M}\right)\)
\(\Rightarrow r=\dfrac{-V}{g}=\dfrac{-\left(-5.12 \times 10^{7}\right)}{6.4}\)
\( = 8 \times {10^6}\;m = 8000\;km\)
\(\therefore\) Height of the point from earth's surface,
\(h=r-R\)
\(\therefore h = 8000 - 6400 = 1600\;km\)
So, correct option is (1).

Explanation:

As all the points on the periphery of either ring are at same distance from point \(P\) the potential at point \(P\) due to whole ring can be calculated as \(V=-(G M)\left(\sqrt{R^{2}+x^{2}}\right)\) where \(x\) is axial distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform are nonuniform.
\(\begin{aligned}& V \text { at } P \text { is } V=-\dfrac{G M}{\sqrt{2 R}}-\dfrac{G 2 M}{\sqrt{5 R}} \\& =-\dfrac{G M}{R}\left[\dfrac{1}{\sqrt{2}}+\dfrac{2}{\sqrt{5}}\right] .\end{aligned}\)

Explanation:

\(V=\dfrac{-G m}{R}-\dfrac{G m}{2 R}=\dfrac{-3 G m}{2 R}\)

Explanation: