02. Gravitational Potential Energy, Gravitational Potential
Gravitation

138473 A satellite moves around the earth in a circular orbit of radius $r$ with speed $v$. If the mass of the satellite is $M$, its total energy is

1 $-1 / 2 \mathrm{Mv}^{2}$
2 $1 / 2 \mathrm{Mv}^{2}$
3 $3 / 2 \mathrm{Mv}^{2}$
4 $\mathrm{Mv}^{2}$
Gravitation

138474 A body of mass $m$ rises to a height $h=\frac{R}{5}$ from
the earth's surface, where $R$ is the earth's radius. If $\mathbf{g}$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1 $\mathrm{mgh}$
2 $\frac{4}{5} \mathrm{mgh}$
3 $\frac{5}{6} \mathrm{mgh}$
4 $\frac{6}{7} \mathrm{mgh}$
Gravitation

138475 The density of a solid sphere of radius $R$ is $\rho(r)=20 \frac{r^{2}}{R^{2}}$ where, $r$ is the distance from its centre. If the gravitational field due to this sphere at a distance $4 R$ from its centre is $E$ and $G$ is the gravitational constant, then the ratio of $\frac{E}{G R}$ is

1 $\frac{\pi}{5}$
2 $3 \pi$
3 $\frac{3 \pi}{2}$
4 $\pi$
Gravitation

138476 From the pole of the earth, a body of mass $m$ is imparted a velocity $v_{0}$ directed vertically up. If $M$ is the mass of the earth, $R$ its radius and $g$ is the free-fall acceleration on its surface, then the height $h$ to which the body will ascent is (neglect air resistance)

1 $\frac{R v_{0}^{2}}{\left(2 g R-v_{0}^{2}\right)}$
2 $\frac{\mathrm{Rv}_{0}^{2}}{2 \mathrm{gR}}$
3 $\mathrm{R}$
4 $\frac{\mathrm{Rv}_{0}^{2}}{\left(2 \mathrm{gR}+\mathrm{v}_{0}^{2}\right)}$
Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $\rho(r)=\rho_{0}\left(\frac{r}{R}\right)^{\beta}$, where $\rho_{0}$ and $\beta$ are constants and $\mathbf{r}$ is the distance from the centre. Let $E_{1}$ and $E_{2}$ be gravitational fields due to sphere at distance $\frac{R}{2}$ and $2 R$ from the centre of sphere. If $\frac{E_{2}}{E_{1}}=4$, the value of $\beta$ is

1 2
2 2.5
3 3
4 4
Gravitation

138473 A satellite moves around the earth in a circular orbit of radius $r$ with speed $v$. If the mass of the satellite is $M$, its total energy is

1 $-1 / 2 \mathrm{Mv}^{2}$
2 $1 / 2 \mathrm{Mv}^{2}$
3 $3 / 2 \mathrm{Mv}^{2}$
4 $\mathrm{Mv}^{2}$
Gravitation

138474 A body of mass $m$ rises to a height $h=\frac{R}{5}$ from
the earth's surface, where $R$ is the earth's radius. If $\mathbf{g}$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1 $\mathrm{mgh}$
2 $\frac{4}{5} \mathrm{mgh}$
3 $\frac{5}{6} \mathrm{mgh}$
4 $\frac{6}{7} \mathrm{mgh}$
Gravitation

138475 The density of a solid sphere of radius $R$ is $\rho(r)=20 \frac{r^{2}}{R^{2}}$ where, $r$ is the distance from its centre. If the gravitational field due to this sphere at a distance $4 R$ from its centre is $E$ and $G$ is the gravitational constant, then the ratio of $\frac{E}{G R}$ is

1 $\frac{\pi}{5}$
2 $3 \pi$
3 $\frac{3 \pi}{2}$
4 $\pi$
Gravitation

138476 From the pole of the earth, a body of mass $m$ is imparted a velocity $v_{0}$ directed vertically up. If $M$ is the mass of the earth, $R$ its radius and $g$ is the free-fall acceleration on its surface, then the height $h$ to which the body will ascent is (neglect air resistance)

1 $\frac{R v_{0}^{2}}{\left(2 g R-v_{0}^{2}\right)}$
2 $\frac{\mathrm{Rv}_{0}^{2}}{2 \mathrm{gR}}$
3 $\mathrm{R}$
4 $\frac{\mathrm{Rv}_{0}^{2}}{\left(2 \mathrm{gR}+\mathrm{v}_{0}^{2}\right)}$
Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $\rho(r)=\rho_{0}\left(\frac{r}{R}\right)^{\beta}$, where $\rho_{0}$ and $\beta$ are constants and $\mathbf{r}$ is the distance from the centre. Let $E_{1}$ and $E_{2}$ be gravitational fields due to sphere at distance $\frac{R}{2}$ and $2 R$ from the centre of sphere. If $\frac{E_{2}}{E_{1}}=4$, the value of $\beta$ is

1 2
2 2.5
3 3
4 4
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Gravitation

138473 A satellite moves around the earth in a circular orbit of radius $r$ with speed $v$. If the mass of the satellite is $M$, its total energy is

1 $-1 / 2 \mathrm{Mv}^{2}$
2 $1 / 2 \mathrm{Mv}^{2}$
3 $3 / 2 \mathrm{Mv}^{2}$
4 $\mathrm{Mv}^{2}$
Gravitation

138474 A body of mass $m$ rises to a height $h=\frac{R}{5}$ from
the earth's surface, where $R$ is the earth's radius. If $\mathbf{g}$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1 $\mathrm{mgh}$
2 $\frac{4}{5} \mathrm{mgh}$
3 $\frac{5}{6} \mathrm{mgh}$
4 $\frac{6}{7} \mathrm{mgh}$
Gravitation

138475 The density of a solid sphere of radius $R$ is $\rho(r)=20 \frac{r^{2}}{R^{2}}$ where, $r$ is the distance from its centre. If the gravitational field due to this sphere at a distance $4 R$ from its centre is $E$ and $G$ is the gravitational constant, then the ratio of $\frac{E}{G R}$ is

1 $\frac{\pi}{5}$
2 $3 \pi$
3 $\frac{3 \pi}{2}$
4 $\pi$
Gravitation

138476 From the pole of the earth, a body of mass $m$ is imparted a velocity $v_{0}$ directed vertically up. If $M$ is the mass of the earth, $R$ its radius and $g$ is the free-fall acceleration on its surface, then the height $h$ to which the body will ascent is (neglect air resistance)

1 $\frac{R v_{0}^{2}}{\left(2 g R-v_{0}^{2}\right)}$
2 $\frac{\mathrm{Rv}_{0}^{2}}{2 \mathrm{gR}}$
3 $\mathrm{R}$
4 $\frac{\mathrm{Rv}_{0}^{2}}{\left(2 \mathrm{gR}+\mathrm{v}_{0}^{2}\right)}$
Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $\rho(r)=\rho_{0}\left(\frac{r}{R}\right)^{\beta}$, where $\rho_{0}$ and $\beta$ are constants and $\mathbf{r}$ is the distance from the centre. Let $E_{1}$ and $E_{2}$ be gravitational fields due to sphere at distance $\frac{R}{2}$ and $2 R$ from the centre of sphere. If $\frac{E_{2}}{E_{1}}=4$, the value of $\beta$ is

1 2
2 2.5
3 3
4 4
Gravitation

138473 A satellite moves around the earth in a circular orbit of radius $r$ with speed $v$. If the mass of the satellite is $M$, its total energy is

1 $-1 / 2 \mathrm{Mv}^{2}$
2 $1 / 2 \mathrm{Mv}^{2}$
3 $3 / 2 \mathrm{Mv}^{2}$
4 $\mathrm{Mv}^{2}$
Gravitation

138474 A body of mass $m$ rises to a height $h=\frac{R}{5}$ from
the earth's surface, where $R$ is the earth's radius. If $\mathbf{g}$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1 $\mathrm{mgh}$
2 $\frac{4}{5} \mathrm{mgh}$
3 $\frac{5}{6} \mathrm{mgh}$
4 $\frac{6}{7} \mathrm{mgh}$
Gravitation

138475 The density of a solid sphere of radius $R$ is $\rho(r)=20 \frac{r^{2}}{R^{2}}$ where, $r$ is the distance from its centre. If the gravitational field due to this sphere at a distance $4 R$ from its centre is $E$ and $G$ is the gravitational constant, then the ratio of $\frac{E}{G R}$ is

1 $\frac{\pi}{5}$
2 $3 \pi$
3 $\frac{3 \pi}{2}$
4 $\pi$
Gravitation

138476 From the pole of the earth, a body of mass $m$ is imparted a velocity $v_{0}$ directed vertically up. If $M$ is the mass of the earth, $R$ its radius and $g$ is the free-fall acceleration on its surface, then the height $h$ to which the body will ascent is (neglect air resistance)

1 $\frac{R v_{0}^{2}}{\left(2 g R-v_{0}^{2}\right)}$
2 $\frac{\mathrm{Rv}_{0}^{2}}{2 \mathrm{gR}}$
3 $\mathrm{R}$
4 $\frac{\mathrm{Rv}_{0}^{2}}{\left(2 \mathrm{gR}+\mathrm{v}_{0}^{2}\right)}$
Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $\rho(r)=\rho_{0}\left(\frac{r}{R}\right)^{\beta}$, where $\rho_{0}$ and $\beta$ are constants and $\mathbf{r}$ is the distance from the centre. Let $E_{1}$ and $E_{2}$ be gravitational fields due to sphere at distance $\frac{R}{2}$ and $2 R$ from the centre of sphere. If $\frac{E_{2}}{E_{1}}=4$, the value of $\beta$ is

1 2
2 2.5
3 3
4 4
Gravitation

138473 A satellite moves around the earth in a circular orbit of radius $r$ with speed $v$. If the mass of the satellite is $M$, its total energy is

1 $-1 / 2 \mathrm{Mv}^{2}$
2 $1 / 2 \mathrm{Mv}^{2}$
3 $3 / 2 \mathrm{Mv}^{2}$
4 $\mathrm{Mv}^{2}$
Gravitation

138474 A body of mass $m$ rises to a height $h=\frac{R}{5}$ from
the earth's surface, where $R$ is the earth's radius. If $\mathbf{g}$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1 $\mathrm{mgh}$
2 $\frac{4}{5} \mathrm{mgh}$
3 $\frac{5}{6} \mathrm{mgh}$
4 $\frac{6}{7} \mathrm{mgh}$
Gravitation

138475 The density of a solid sphere of radius $R$ is $\rho(r)=20 \frac{r^{2}}{R^{2}}$ where, $r$ is the distance from its centre. If the gravitational field due to this sphere at a distance $4 R$ from its centre is $E$ and $G$ is the gravitational constant, then the ratio of $\frac{E}{G R}$ is

1 $\frac{\pi}{5}$
2 $3 \pi$
3 $\frac{3 \pi}{2}$
4 $\pi$
Gravitation

138476 From the pole of the earth, a body of mass $m$ is imparted a velocity $v_{0}$ directed vertically up. If $M$ is the mass of the earth, $R$ its radius and $g$ is the free-fall acceleration on its surface, then the height $h$ to which the body will ascent is (neglect air resistance)

1 $\frac{R v_{0}^{2}}{\left(2 g R-v_{0}^{2}\right)}$
2 $\frac{\mathrm{Rv}_{0}^{2}}{2 \mathrm{gR}}$
3 $\mathrm{R}$
4 $\frac{\mathrm{Rv}_{0}^{2}}{\left(2 \mathrm{gR}+\mathrm{v}_{0}^{2}\right)}$
Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $\rho(r)=\rho_{0}\left(\frac{r}{R}\right)^{\beta}$, where $\rho_{0}$ and $\beta$ are constants and $\mathbf{r}$ is the distance from the centre. Let $E_{1}$ and $E_{2}$ be gravitational fields due to sphere at distance $\frac{R}{2}$ and $2 R$ from the centre of sphere. If $\frac{E_{2}}{E_{1}}=4$, the value of $\beta$ is

1 2
2 2.5
3 3
4 4