138458 A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_{e}\right)$, the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth)

138459 An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth' surface, ignoring resistance, will be

138460 If the gravitational filed in the space is given as $\left(-\frac{K}{r^{2}}\right)$ Taking the reference point to be at $\mathbf{r}=\mathbf{2}$ cm with gravitational potential $V=10 \mathrm{~J} / \mathrm{kg}$. Find the gravitational potential at $r=3 \mathrm{~cm}$ in SI unit (Given that, $K=6 \mathrm{Jcm} / \mathrm{kg}$ )

138461 A body of mass $m$ is released from a height equal to the radius $R$ of the Earth. The velocity with which it will strike the Earth's surface is

138463 Infinite number of masses, each $1 \mathrm{~kg}$, are placed along the $x$-axis at $x= \pm 1 \mathrm{~m} . \pm 2 \mathrm{~m} . \pm 4$ $\mathrm{m}, \pm 8 \mathrm{~m}, \pm 16 \mathrm{~m} \ldots .$. The magnitude of the resultant gravitational potential in terms of gravitational constant $G$ at the origin $(x=0)$ is

138458 A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_{e}\right)$, the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth)

138459 An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth' surface, ignoring resistance, will be

138460 If the gravitational filed in the space is given as $\left(-\frac{K}{r^{2}}\right)$ Taking the reference point to be at $\mathbf{r}=\mathbf{2}$ cm with gravitational potential $V=10 \mathrm{~J} / \mathrm{kg}$. Find the gravitational potential at $r=3 \mathrm{~cm}$ in SI unit (Given that, $K=6 \mathrm{Jcm} / \mathrm{kg}$ )

138461 A body of mass $m$ is released from a height equal to the radius $R$ of the Earth. The velocity with which it will strike the Earth's surface is

138463 Infinite number of masses, each $1 \mathrm{~kg}$, are placed along the $x$-axis at $x= \pm 1 \mathrm{~m} . \pm 2 \mathrm{~m} . \pm 4$ $\mathrm{m}, \pm 8 \mathrm{~m}, \pm 16 \mathrm{~m} \ldots .$. The magnitude of the resultant gravitational potential in terms of gravitational constant $G$ at the origin $(x=0)$ is

138458 A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_{e}\right)$, the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth)

138459 An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth' surface, ignoring resistance, will be

138460 If the gravitational filed in the space is given as $\left(-\frac{K}{r^{2}}\right)$ Taking the reference point to be at $\mathbf{r}=\mathbf{2}$ cm with gravitational potential $V=10 \mathrm{~J} / \mathrm{kg}$. Find the gravitational potential at $r=3 \mathrm{~cm}$ in SI unit (Given that, $K=6 \mathrm{Jcm} / \mathrm{kg}$ )

138461 A body of mass $m$ is released from a height equal to the radius $R$ of the Earth. The velocity with which it will strike the Earth's surface is

138463 Infinite number of masses, each $1 \mathrm{~kg}$, are placed along the $x$-axis at $x= \pm 1 \mathrm{~m} . \pm 2 \mathrm{~m} . \pm 4$ $\mathrm{m}, \pm 8 \mathrm{~m}, \pm 16 \mathrm{~m} \ldots .$. The magnitude of the resultant gravitational potential in terms of gravitational constant $G$ at the origin $(x=0)$ is

138458 A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_{e}\right)$, the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth)

138459 An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth' surface, ignoring resistance, will be

138460 If the gravitational filed in the space is given as $\left(-\frac{K}{r^{2}}\right)$ Taking the reference point to be at $\mathbf{r}=\mathbf{2}$ cm with gravitational potential $V=10 \mathrm{~J} / \mathrm{kg}$. Find the gravitational potential at $r=3 \mathrm{~cm}$ in SI unit (Given that, $K=6 \mathrm{Jcm} / \mathrm{kg}$ )

138461 A body of mass $m$ is released from a height equal to the radius $R$ of the Earth. The velocity with which it will strike the Earth's surface is

138463 Infinite number of masses, each $1 \mathrm{~kg}$, are placed along the $x$-axis at $x= \pm 1 \mathrm{~m} . \pm 2 \mathrm{~m} . \pm 4$ $\mathrm{m}, \pm 8 \mathrm{~m}, \pm 16 \mathrm{~m} \ldots .$. The magnitude of the resultant gravitational potential in terms of gravitational constant $G$ at the origin $(x=0)$ is

138458 A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_{e}\right)$, the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth)

138459 An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth' surface, ignoring resistance, will be

138460 If the gravitational filed in the space is given as $\left(-\frac{K}{r^{2}}\right)$ Taking the reference point to be at $\mathbf{r}=\mathbf{2}$ cm with gravitational potential $V=10 \mathrm{~J} / \mathrm{kg}$. Find the gravitational potential at $r=3 \mathrm{~cm}$ in SI unit (Given that, $K=6 \mathrm{Jcm} / \mathrm{kg}$ )

138461 A body of mass $m$ is released from a height equal to the radius $R$ of the Earth. The velocity with which it will strike the Earth's surface is

138463 Infinite number of masses, each $1 \mathrm{~kg}$, are placed along the $x$-axis at $x= \pm 1 \mathrm{~m} . \pm 2 \mathrm{~m} . \pm 4$ $\mathrm{m}, \pm 8 \mathrm{~m}, \pm 16 \mathrm{~m} \ldots .$. The magnitude of the resultant gravitational potential in terms of gravitational constant $G$ at the origin $(x=0)$ is