NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138468
There is a shell of mass $M$ and density of the shell is uniform. The work done to take a point mass from point $A$ to $B$ is $[A B=r]$
1 $\frac{\mathrm{GmM}}{\mathrm{r}}$
2 $\frac{\mathrm{GmM}}{\mathrm{R}}$
3 $-\frac{\mathrm{GmM}}{\mathrm{r}}$
4 zero
Explanation:
D As we know, gravitational field inside the shell is zero. Hence, there is no work required.
BITSAT-2013
Gravitation
138469
Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $r$. The gravitational potential at a point on the line joining them, where the gravitational field is zero, is
1 zero
2 $-\frac{4 \mathrm{Gm}}{\mathrm{r}}$
3 $-\frac{6 \mathrm{Gm}}{\mathrm{r}}$
4 $-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
Explanation:
D Two masses $\mathrm{m}$ and $4 \mathrm{~m}$ kept at distance r.Let gravitational field be zero at a distance y from mass $\mathrm{m}$ at a point $\mathrm{C}$. $\frac{\mathrm{Gm}}{\mathrm{y}^{2}}=\frac{\mathrm{G} \times 4 \mathrm{~m}}{(\mathrm{r}-\mathrm{y})^{2}}$ $(2 \mathrm{y})^{2}=(\mathrm{r}-\mathrm{y})^{2}$ $\Rightarrow 2 \mathrm{y}=\mathrm{r}-\mathrm{y}$ $\Rightarrow \mathrm{y}=\frac{\mathrm{r}}{3}$ Potential at point $\mathrm{C}$, $\mathrm{V}_{\mathrm{C}} =-\frac{\mathrm{Gm}}{\mathrm{y}}-\frac{\mathrm{G}(4 \mathrm{~m})}{\mathrm{r}-\mathrm{y}}$ $=-\frac{3 \mathrm{Gm}}{\mathrm{r}}-\frac{6 \mathrm{Gm}}{\mathrm{r}}$ $\therefore \quad =-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
CG PET -2018
Gravitation
138470
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $R / 2$ is removed, as shown in figure. Taking gravitational potential $V=0$ at $r=\alpha$. The potential at the centre of the cavity thus formed is $(G=$ gravitational constant)
1 $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$
2 $\frac{-2 \mathrm{GM}}{\mathrm{R}}$
3 $\frac{-\mathrm{GM}}{2 \mathrm{R}}$
4 $\frac{-\mathrm{GM}}{\mathrm{R}}$
Explanation:
D Given, Gravitational potential $(\mathrm{V})=0$ Distance $(\mathrm{r})=\alpha$ Mass of sphere $=\mathrm{M}$ Radius of sphere $=\mathrm{R}$ Spherical portion removed has radius $=\frac{\mathrm{R}}{2}$ Gravitational potential at internal point of solid sphere at a distance $r, V=-\frac{G M}{R}\left[\frac{3}{2}-\frac{r^{2}}{2 R^{2}}\right]$ When $\mathrm{r}=\frac{\mathrm{R}}{2}$, $\mathrm{V}_{1} =-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{2}-\frac{\mathrm{R}^{2}}{8 \mathrm{R}^{2}}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{12-1}{8}\right]$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ When radius of spherical portion is removed, $\mathrm{V}_{2} =-\frac{3}{2} \frac{\mathrm{GM} / 8}{\mathrm{R} / 2}=-\frac{3}{2} \times \frac{2 \mathrm{GM}}{8 \mathrm{R}}$ $=-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ Net potential at the centre of the gravity $(\mathrm{V})=\mathrm{V}_{1}-\mathrm{V}_{2}$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}\right)$ $= -\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}+\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ $=\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{8}-\frac{11}{8}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}$
CG PET -2016
Gravitation
138472
If $g$ be the acceleration due to gravity at the earth surface, then what will be the increase in potential energy if object of mass $m$ is raised by its radius $R$ ?
1 $\frac{1}{2} \mathrm{mgR}$
2 $2 \mathrm{mgR}$
3 $\mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
A Given, Acceleration due to gravity at the earth surface $=g$ Mass of object $=\mathrm{m}$ Radius of earth $=\mathrm{R}$ Mass of earth $=\mathrm{M}$ Gravitational potential energy of body on the earth's surface $\left(\mathrm{U}_{1}\right)=-\frac{\mathrm{GMm}}{\mathrm{R}}$ Potential energy of the object at a height $(\mathrm{h})=\mathrm{R}$ from the earth surface $\left(\mathrm{U}_{2}\right)=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{R})} \quad[\because \mathrm{h}=\mathrm{R}]$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(2 \mathrm{R})}$ Increases potential energy $(\mathrm{U})=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$ $\mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ Multiplying $\mathrm{R}$ in numerator and denominator, $\mathrm{U}=\frac{\mathrm{GMmR}}{2 \mathrm{R}^{2}}$ So, $\mathrm{U}=\frac{\mathrm{gmR}}{2} \quad\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\mathrm{U}=\frac{1}{2} \mathrm{mgR}$
138468
There is a shell of mass $M$ and density of the shell is uniform. The work done to take a point mass from point $A$ to $B$ is $[A B=r]$
1 $\frac{\mathrm{GmM}}{\mathrm{r}}$
2 $\frac{\mathrm{GmM}}{\mathrm{R}}$
3 $-\frac{\mathrm{GmM}}{\mathrm{r}}$
4 zero
Explanation:
D As we know, gravitational field inside the shell is zero. Hence, there is no work required.
BITSAT-2013
Gravitation
138469
Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $r$. The gravitational potential at a point on the line joining them, where the gravitational field is zero, is
1 zero
2 $-\frac{4 \mathrm{Gm}}{\mathrm{r}}$
3 $-\frac{6 \mathrm{Gm}}{\mathrm{r}}$
4 $-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
Explanation:
D Two masses $\mathrm{m}$ and $4 \mathrm{~m}$ kept at distance r.Let gravitational field be zero at a distance y from mass $\mathrm{m}$ at a point $\mathrm{C}$. $\frac{\mathrm{Gm}}{\mathrm{y}^{2}}=\frac{\mathrm{G} \times 4 \mathrm{~m}}{(\mathrm{r}-\mathrm{y})^{2}}$ $(2 \mathrm{y})^{2}=(\mathrm{r}-\mathrm{y})^{2}$ $\Rightarrow 2 \mathrm{y}=\mathrm{r}-\mathrm{y}$ $\Rightarrow \mathrm{y}=\frac{\mathrm{r}}{3}$ Potential at point $\mathrm{C}$, $\mathrm{V}_{\mathrm{C}} =-\frac{\mathrm{Gm}}{\mathrm{y}}-\frac{\mathrm{G}(4 \mathrm{~m})}{\mathrm{r}-\mathrm{y}}$ $=-\frac{3 \mathrm{Gm}}{\mathrm{r}}-\frac{6 \mathrm{Gm}}{\mathrm{r}}$ $\therefore \quad =-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
CG PET -2018
Gravitation
138470
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $R / 2$ is removed, as shown in figure. Taking gravitational potential $V=0$ at $r=\alpha$. The potential at the centre of the cavity thus formed is $(G=$ gravitational constant)
1 $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$
2 $\frac{-2 \mathrm{GM}}{\mathrm{R}}$
3 $\frac{-\mathrm{GM}}{2 \mathrm{R}}$
4 $\frac{-\mathrm{GM}}{\mathrm{R}}$
Explanation:
D Given, Gravitational potential $(\mathrm{V})=0$ Distance $(\mathrm{r})=\alpha$ Mass of sphere $=\mathrm{M}$ Radius of sphere $=\mathrm{R}$ Spherical portion removed has radius $=\frac{\mathrm{R}}{2}$ Gravitational potential at internal point of solid sphere at a distance $r, V=-\frac{G M}{R}\left[\frac{3}{2}-\frac{r^{2}}{2 R^{2}}\right]$ When $\mathrm{r}=\frac{\mathrm{R}}{2}$, $\mathrm{V}_{1} =-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{2}-\frac{\mathrm{R}^{2}}{8 \mathrm{R}^{2}}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{12-1}{8}\right]$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ When radius of spherical portion is removed, $\mathrm{V}_{2} =-\frac{3}{2} \frac{\mathrm{GM} / 8}{\mathrm{R} / 2}=-\frac{3}{2} \times \frac{2 \mathrm{GM}}{8 \mathrm{R}}$ $=-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ Net potential at the centre of the gravity $(\mathrm{V})=\mathrm{V}_{1}-\mathrm{V}_{2}$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}\right)$ $= -\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}+\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ $=\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{8}-\frac{11}{8}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}$
CG PET -2016
Gravitation
138472
If $g$ be the acceleration due to gravity at the earth surface, then what will be the increase in potential energy if object of mass $m$ is raised by its radius $R$ ?
1 $\frac{1}{2} \mathrm{mgR}$
2 $2 \mathrm{mgR}$
3 $\mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
A Given, Acceleration due to gravity at the earth surface $=g$ Mass of object $=\mathrm{m}$ Radius of earth $=\mathrm{R}$ Mass of earth $=\mathrm{M}$ Gravitational potential energy of body on the earth's surface $\left(\mathrm{U}_{1}\right)=-\frac{\mathrm{GMm}}{\mathrm{R}}$ Potential energy of the object at a height $(\mathrm{h})=\mathrm{R}$ from the earth surface $\left(\mathrm{U}_{2}\right)=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{R})} \quad[\because \mathrm{h}=\mathrm{R}]$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(2 \mathrm{R})}$ Increases potential energy $(\mathrm{U})=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$ $\mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ Multiplying $\mathrm{R}$ in numerator and denominator, $\mathrm{U}=\frac{\mathrm{GMmR}}{2 \mathrm{R}^{2}}$ So, $\mathrm{U}=\frac{\mathrm{gmR}}{2} \quad\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\mathrm{U}=\frac{1}{2} \mathrm{mgR}$
138468
There is a shell of mass $M$ and density of the shell is uniform. The work done to take a point mass from point $A$ to $B$ is $[A B=r]$
1 $\frac{\mathrm{GmM}}{\mathrm{r}}$
2 $\frac{\mathrm{GmM}}{\mathrm{R}}$
3 $-\frac{\mathrm{GmM}}{\mathrm{r}}$
4 zero
Explanation:
D As we know, gravitational field inside the shell is zero. Hence, there is no work required.
BITSAT-2013
Gravitation
138469
Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $r$. The gravitational potential at a point on the line joining them, where the gravitational field is zero, is
1 zero
2 $-\frac{4 \mathrm{Gm}}{\mathrm{r}}$
3 $-\frac{6 \mathrm{Gm}}{\mathrm{r}}$
4 $-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
Explanation:
D Two masses $\mathrm{m}$ and $4 \mathrm{~m}$ kept at distance r.Let gravitational field be zero at a distance y from mass $\mathrm{m}$ at a point $\mathrm{C}$. $\frac{\mathrm{Gm}}{\mathrm{y}^{2}}=\frac{\mathrm{G} \times 4 \mathrm{~m}}{(\mathrm{r}-\mathrm{y})^{2}}$ $(2 \mathrm{y})^{2}=(\mathrm{r}-\mathrm{y})^{2}$ $\Rightarrow 2 \mathrm{y}=\mathrm{r}-\mathrm{y}$ $\Rightarrow \mathrm{y}=\frac{\mathrm{r}}{3}$ Potential at point $\mathrm{C}$, $\mathrm{V}_{\mathrm{C}} =-\frac{\mathrm{Gm}}{\mathrm{y}}-\frac{\mathrm{G}(4 \mathrm{~m})}{\mathrm{r}-\mathrm{y}}$ $=-\frac{3 \mathrm{Gm}}{\mathrm{r}}-\frac{6 \mathrm{Gm}}{\mathrm{r}}$ $\therefore \quad =-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
CG PET -2018
Gravitation
138470
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $R / 2$ is removed, as shown in figure. Taking gravitational potential $V=0$ at $r=\alpha$. The potential at the centre of the cavity thus formed is $(G=$ gravitational constant)
1 $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$
2 $\frac{-2 \mathrm{GM}}{\mathrm{R}}$
3 $\frac{-\mathrm{GM}}{2 \mathrm{R}}$
4 $\frac{-\mathrm{GM}}{\mathrm{R}}$
Explanation:
D Given, Gravitational potential $(\mathrm{V})=0$ Distance $(\mathrm{r})=\alpha$ Mass of sphere $=\mathrm{M}$ Radius of sphere $=\mathrm{R}$ Spherical portion removed has radius $=\frac{\mathrm{R}}{2}$ Gravitational potential at internal point of solid sphere at a distance $r, V=-\frac{G M}{R}\left[\frac{3}{2}-\frac{r^{2}}{2 R^{2}}\right]$ When $\mathrm{r}=\frac{\mathrm{R}}{2}$, $\mathrm{V}_{1} =-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{2}-\frac{\mathrm{R}^{2}}{8 \mathrm{R}^{2}}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{12-1}{8}\right]$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ When radius of spherical portion is removed, $\mathrm{V}_{2} =-\frac{3}{2} \frac{\mathrm{GM} / 8}{\mathrm{R} / 2}=-\frac{3}{2} \times \frac{2 \mathrm{GM}}{8 \mathrm{R}}$ $=-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ Net potential at the centre of the gravity $(\mathrm{V})=\mathrm{V}_{1}-\mathrm{V}_{2}$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}\right)$ $= -\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}+\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ $=\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{8}-\frac{11}{8}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}$
CG PET -2016
Gravitation
138472
If $g$ be the acceleration due to gravity at the earth surface, then what will be the increase in potential energy if object of mass $m$ is raised by its radius $R$ ?
1 $\frac{1}{2} \mathrm{mgR}$
2 $2 \mathrm{mgR}$
3 $\mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
A Given, Acceleration due to gravity at the earth surface $=g$ Mass of object $=\mathrm{m}$ Radius of earth $=\mathrm{R}$ Mass of earth $=\mathrm{M}$ Gravitational potential energy of body on the earth's surface $\left(\mathrm{U}_{1}\right)=-\frac{\mathrm{GMm}}{\mathrm{R}}$ Potential energy of the object at a height $(\mathrm{h})=\mathrm{R}$ from the earth surface $\left(\mathrm{U}_{2}\right)=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{R})} \quad[\because \mathrm{h}=\mathrm{R}]$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(2 \mathrm{R})}$ Increases potential energy $(\mathrm{U})=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$ $\mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ Multiplying $\mathrm{R}$ in numerator and denominator, $\mathrm{U}=\frac{\mathrm{GMmR}}{2 \mathrm{R}^{2}}$ So, $\mathrm{U}=\frac{\mathrm{gmR}}{2} \quad\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\mathrm{U}=\frac{1}{2} \mathrm{mgR}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138468
There is a shell of mass $M$ and density of the shell is uniform. The work done to take a point mass from point $A$ to $B$ is $[A B=r]$
1 $\frac{\mathrm{GmM}}{\mathrm{r}}$
2 $\frac{\mathrm{GmM}}{\mathrm{R}}$
3 $-\frac{\mathrm{GmM}}{\mathrm{r}}$
4 zero
Explanation:
D As we know, gravitational field inside the shell is zero. Hence, there is no work required.
BITSAT-2013
Gravitation
138469
Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $r$. The gravitational potential at a point on the line joining them, where the gravitational field is zero, is
1 zero
2 $-\frac{4 \mathrm{Gm}}{\mathrm{r}}$
3 $-\frac{6 \mathrm{Gm}}{\mathrm{r}}$
4 $-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
Explanation:
D Two masses $\mathrm{m}$ and $4 \mathrm{~m}$ kept at distance r.Let gravitational field be zero at a distance y from mass $\mathrm{m}$ at a point $\mathrm{C}$. $\frac{\mathrm{Gm}}{\mathrm{y}^{2}}=\frac{\mathrm{G} \times 4 \mathrm{~m}}{(\mathrm{r}-\mathrm{y})^{2}}$ $(2 \mathrm{y})^{2}=(\mathrm{r}-\mathrm{y})^{2}$ $\Rightarrow 2 \mathrm{y}=\mathrm{r}-\mathrm{y}$ $\Rightarrow \mathrm{y}=\frac{\mathrm{r}}{3}$ Potential at point $\mathrm{C}$, $\mathrm{V}_{\mathrm{C}} =-\frac{\mathrm{Gm}}{\mathrm{y}}-\frac{\mathrm{G}(4 \mathrm{~m})}{\mathrm{r}-\mathrm{y}}$ $=-\frac{3 \mathrm{Gm}}{\mathrm{r}}-\frac{6 \mathrm{Gm}}{\mathrm{r}}$ $\therefore \quad =-\frac{9 \mathrm{Gm}}{\mathrm{r}}$
CG PET -2018
Gravitation
138470
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $R / 2$ is removed, as shown in figure. Taking gravitational potential $V=0$ at $r=\alpha$. The potential at the centre of the cavity thus formed is $(G=$ gravitational constant)
1 $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$
2 $\frac{-2 \mathrm{GM}}{\mathrm{R}}$
3 $\frac{-\mathrm{GM}}{2 \mathrm{R}}$
4 $\frac{-\mathrm{GM}}{\mathrm{R}}$
Explanation:
D Given, Gravitational potential $(\mathrm{V})=0$ Distance $(\mathrm{r})=\alpha$ Mass of sphere $=\mathrm{M}$ Radius of sphere $=\mathrm{R}$ Spherical portion removed has radius $=\frac{\mathrm{R}}{2}$ Gravitational potential at internal point of solid sphere at a distance $r, V=-\frac{G M}{R}\left[\frac{3}{2}-\frac{r^{2}}{2 R^{2}}\right]$ When $\mathrm{r}=\frac{\mathrm{R}}{2}$, $\mathrm{V}_{1} =-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{2}-\frac{\mathrm{R}^{2}}{8 \mathrm{R}^{2}}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{12-1}{8}\right]$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ When radius of spherical portion is removed, $\mathrm{V}_{2} =-\frac{3}{2} \frac{\mathrm{GM} / 8}{\mathrm{R} / 2}=-\frac{3}{2} \times \frac{2 \mathrm{GM}}{8 \mathrm{R}}$ $=-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ Net potential at the centre of the gravity $(\mathrm{V})=\mathrm{V}_{1}-\mathrm{V}_{2}$ $=-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}\right)$ $= -\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}}+\frac{3}{8} \frac{\mathrm{GM}}{\mathrm{R}}$ $=\frac{\mathrm{GM}}{\mathrm{R}}\left[\frac{3}{8}-\frac{11}{8}\right]$ $=-\frac{\mathrm{GM}}{\mathrm{R}}$
CG PET -2016
Gravitation
138472
If $g$ be the acceleration due to gravity at the earth surface, then what will be the increase in potential energy if object of mass $m$ is raised by its radius $R$ ?
1 $\frac{1}{2} \mathrm{mgR}$
2 $2 \mathrm{mgR}$
3 $\mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
A Given, Acceleration due to gravity at the earth surface $=g$ Mass of object $=\mathrm{m}$ Radius of earth $=\mathrm{R}$ Mass of earth $=\mathrm{M}$ Gravitational potential energy of body on the earth's surface $\left(\mathrm{U}_{1}\right)=-\frac{\mathrm{GMm}}{\mathrm{R}}$ Potential energy of the object at a height $(\mathrm{h})=\mathrm{R}$ from the earth surface $\left(\mathrm{U}_{2}\right)=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{R})} \quad[\because \mathrm{h}=\mathrm{R}]$ $\mathrm{U}_{2} =-\frac{\mathrm{GMm}}{(2 \mathrm{R})}$ Increases potential energy $(\mathrm{U})=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\mathrm{U}=-\frac{\mathrm{GMm}}{2 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$ $\mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ Multiplying $\mathrm{R}$ in numerator and denominator, $\mathrm{U}=\frac{\mathrm{GMmR}}{2 \mathrm{R}^{2}}$ So, $\mathrm{U}=\frac{\mathrm{gmR}}{2} \quad\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\mathrm{U}=\frac{1}{2} \mathrm{mgR}$
