QBManager

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 {Mv}^{2}

2

1 / 2 {Mv}^{2}

3

3 / 2 {Mv}^{2}

4

{Mv}^{2}

Explanation:

A Given,
Speed of satellite $= v$
Radius of circular orbit of satellite $= r$
Mass of satellite $= M$
Let Mass of earth $= M_{e}$
Radius of earth $= R_{e}$
We know,
Kinetic energy of Satellite (K.E.) $= \frac{1}{2} {Mv}^{2}$
And Orbital velocity of satellite $(v) = \sqrt{{gR}_{e}}$
Squaring both sides,
$v^{2} = {gR}_{e}$
Then,
$K.E. = \frac{1}{2} Mg R_{e}$
Potential energy of a satellite
$P.E. = - \frac{{GM}_{e} M}{R_{e}}$
In numerator and denominator multiply by $R_{e}$ ,
$P.E. = - \frac{{GM}_{e} M \cdot R_{e}}{R_{e}^{2}}$
Then,
$P.E. = - {gMR}_{e} (∵ g = \frac{{GM}_{e}}{R_{e}^{2}})$
Now,
Total Energy $=$ K.E. + P.E.
$= \frac{1}{2} {MgR}_{e} - {MgR}_{e}$
$= - \frac{1}{2} {MgR}_{e} [∵ v^{2} = {gR}_{e}]$
$= - \frac{1}{2} {Mv}^{2}$

MP PMT 2001

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 $g$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1

mgh

2

\frac{4}{5} mgh

3

\frac{5}{6} mgh

4

\frac{6}{7} mgh

Explanation:

C Given, Mass of earth $= M$
Mass of body $= m$
Radius of earth $= R$
We know,
Potential Energy on the earth's surface (P.E. $)_{1} = - \frac{G M m}{R}$
And Potential Energy at height $(h) = \frac{R}{5}$ from the earth's surface $(P.E.)_{2} = - \frac{GMm}{(R + \frac{R}{5})}$
$(P.E.)_{2} = - \frac{GMm}{(\frac{5 R + R}{5})}$
$(P.E.)_{2} = \frac{- 5 GMm}{6 R}$
Increase in potential energy (P.E. $) = (P.E.)_{2} - (P.E.)_{1}$
$P.E. = - \frac{5 GMm}{6 R} - (- \frac{GMm}{R})$
$P.E. = - \frac{5 GMm}{6 R} + \frac{GMm}{R}$
$P.E. = \frac{- 5 GMm + 6 GMm}{6 R}$
$P.E. = \frac{GMm}{6 R}$
Multiplying $R$ in numerator and denominator,
$P.E. = \frac{GMmR}{6 R^{2}}$
Then,
$\begin{array}{lr} P.E. = \frac{gmR}{6} & (∵ g = \frac{GM}{R^{2}}) \\ P.E. = \frac{gm (5 h)}{6} & [\begin{array}{l} ∵ h = \frac{R}{5} \\ R = 5 h \end{array}] \end{array}$
Hence, P.E. $= \frac{5}{6} mgh$

Manipal UGET -2020

Gravitation

138475 The density of a solid sphere of radius $R$ is $ρ (r) = 20 \frac{r^{2}}{R^{2}}$ 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{π}{5}

2

3 π

3

\frac{3 π}{2}

4

π

Explanation:

D Given,
Density of a solid sphere $[ρ (r)] = 20 \frac{r^{2}}{R^{2}}$
Distance from centre $= r$
We know,
Volume of a spherical shell of radius $r$ and thickness $dr$ is,
$dV = 4 π r^{2} dr$
And, Density of a solid sphere $[ρ (r)] = \frac{dM}{dV}$
Then,
$dM = \frac{20 r^{2}}{R^{2}} (4 π r^{2} dr)$
$dM = \frac{80 π r^{4}}{R^{2}} dr$
$Mass of the shell (dM) = ρ . d V$
Total mass of complete solid sphere, $M = \int_{0}^{R} d M$
$M = \frac{80 π}{R^{2}} \int_{0}^{R} r^{4} dr$
$M = \frac{80 π}{R^{2}} {[\frac{r^{5}}{5}]}_{0}^{R}$
$M = \frac{80 π}{R^{2}} \times \frac{1}{5} [R^{5} - (0)^{5}]$
$M = \frac{16 π}{R^{2}} R^{5}$
$∴ M = 16 π R^{3}$
Gravitational field intensity at a distance $4 R$ from its centre $(E) = \frac{G M}{(4 R)^{2}}$
Then,
$E = \frac{G (16 π R^{3})}{16 R^{2}}$
$\Rightarrow E = G π R$
$∴ \frac{E}{GR} = π$

TS- EAMCET-04.05.2018

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}}{(2 g R - v_{0}^{2})}

2

\frac{{Rv}_{0}^{2}}{2 gR}

3

R

4

\frac{{Rv}_{0}^{2}}{(2 gR + v_{0}^{2})}

Explanation:

A Given,
Mass of body $= m$
Velocity of body $= v_{0}$
Mass of the earth $= M$
Radius of the earth $= R$
Free-fall acceleration on its surface $= g$ Height $= h$
We know,
Kinetic energy of the body (K.E. $)_{initial} = \frac{1}{2} {mv}_{0}^{2}$
And Potential energy of the body $(U)_{initial} = - \frac{G M m}{R}$ Now,
Potential energy of the body at height $h$ $(U)_{final} = - \frac{GMm}{(R + h)}$
According to law of conservation of energy,
$(K.E.)_{initial} + (U)_{initial} = (U)_{final}$
$\frac{1}{2} {mv}_{0}^{2} + (- \frac{GMm}{R}) = - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = \frac{GMm}{R} - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{1}{R} - \frac{1}{(R + h)})$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{R + h - R}{R (R + h)})$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R (R + h)}$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R^{2} (1 + \frac{h}{R})}$
Then,
$\frac{v_{0}^{2}}{2} = \frac{gh}{(1 + \frac{h}{R})} (∵ g = \frac{GM}{R^{2}})$
$\frac{v_{0}^{2}}{2} = \frac{ghR}{(R + h)}$
Now,
$v_{0}^{2} \times (R + h) = 2 g R h$
$v_{0}^{2} R + v_{0}^{2} h = 2 g R h$
$2 g R h - v_{0}^{2} h = R v_{0}^{2}$
$h (2 g R - v_{0}^{2}) = R v_{0}^{2}$
Hence, $h = \frac{{Rv}_{0}^{2}}{2 gR - v_{0}^{2}}$

MHT CET 2020

Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $ρ (r) = ρ_{0} {(\frac{r}{R})}^{β}$ , where $ρ_{0}$ and $β$ are constants and $r$ 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 $β$ is

1 2

2 2.5

3 3

4 4

Explanation:

C
Mass enclosed in a sphere of radius $r, M_{r} = \int ρ d V$
$= \int ρ_{0} {(\frac{r}{R})}^{β} 4 π r^{2} d r$
$= {[\frac{4 π ρ_{0}}{R^{β}} \times \frac{r^{β + 3}}{β + 3}]}_{0}^{r}$
When $radius (r) = \frac{R}{2}$ ,
Mass enclosed in a sphere $M_{1} = \frac{4 π ρ_{0}}{R^{β}} \frac{{(\frac{R}{2})}^{β + 3}}{(β + 3)}$
$= \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
Gravitational field intensity $E_{1} = \frac{{GM}_{1}}{{(\frac{R}{2})}^{2}}$
$= \frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
When radius $(r) = R$ ,
Mass enclosed in a sphere,
$M_{2} = \frac{4 π ρ_{0}}{R^{β}} \cdot \frac{R^{β + 3}}{β + 3} = \frac{4 π ρ_{0} R^{3}}{β + 3}$
Gravitational field intensity $(E_{2}) = \frac{{GM}_{2}}{(2 R)^{2}}$
$= \frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{β + 3}$
If $\frac{E_{2}}{E_{1}} = 4$ ,
$\frac{\frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)}}{\frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}} = 4$
Now,
$\frac{2^{β + 3}}{16} = 4$
$2^{β + 3} = 64$
$2^{β} \cdot 2^{3} = 2^{6}$
$2^{β} = 2^{3}$
$β = 3$

TS- EAMCET-05.05.2018

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 {Mv}^{2}

2

1 / 2 {Mv}^{2}

3

3 / 2 {Mv}^{2}

4

{Mv}^{2}

Explanation:

A Given,
Speed of satellite $= v$
Radius of circular orbit of satellite $= r$
Mass of satellite $= M$
Let Mass of earth $= M_{e}$
Radius of earth $= R_{e}$
We know,
Kinetic energy of Satellite (K.E.) $= \frac{1}{2} {Mv}^{2}$
And Orbital velocity of satellite $(v) = \sqrt{{gR}_{e}}$
Squaring both sides,
$v^{2} = {gR}_{e}$
Then,
$K.E. = \frac{1}{2} Mg R_{e}$
Potential energy of a satellite
$P.E. = - \frac{{GM}_{e} M}{R_{e}}$
In numerator and denominator multiply by $R_{e}$ ,
$P.E. = - \frac{{GM}_{e} M \cdot R_{e}}{R_{e}^{2}}$
Then,
$P.E. = - {gMR}_{e} (∵ g = \frac{{GM}_{e}}{R_{e}^{2}})$
Now,
Total Energy $=$ K.E. + P.E.
$= \frac{1}{2} {MgR}_{e} - {MgR}_{e}$
$= - \frac{1}{2} {MgR}_{e} [∵ v^{2} = {gR}_{e}]$
$= - \frac{1}{2} {Mv}^{2}$

MP PMT 2001

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 $g$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1

mgh

2

\frac{4}{5} mgh

3

\frac{5}{6} mgh

4

\frac{6}{7} mgh

Explanation:

C Given, Mass of earth $= M$
Mass of body $= m$
Radius of earth $= R$
We know,
Potential Energy on the earth's surface (P.E. $)_{1} = - \frac{G M m}{R}$
And Potential Energy at height $(h) = \frac{R}{5}$ from the earth's surface $(P.E.)_{2} = - \frac{GMm}{(R + \frac{R}{5})}$
$(P.E.)_{2} = - \frac{GMm}{(\frac{5 R + R}{5})}$
$(P.E.)_{2} = \frac{- 5 GMm}{6 R}$
Increase in potential energy (P.E. $) = (P.E.)_{2} - (P.E.)_{1}$
$P.E. = - \frac{5 GMm}{6 R} - (- \frac{GMm}{R})$
$P.E. = - \frac{5 GMm}{6 R} + \frac{GMm}{R}$
$P.E. = \frac{- 5 GMm + 6 GMm}{6 R}$
$P.E. = \frac{GMm}{6 R}$
Multiplying $R$ in numerator and denominator,
$P.E. = \frac{GMmR}{6 R^{2}}$
Then,
$\begin{array}{lr} P.E. = \frac{gmR}{6} & (∵ g = \frac{GM}{R^{2}}) \\ P.E. = \frac{gm (5 h)}{6} & [\begin{array}{l} ∵ h = \frac{R}{5} \\ R = 5 h \end{array}] \end{array}$
Hence, P.E. $= \frac{5}{6} mgh$

Manipal UGET -2020

Gravitation

138475 The density of a solid sphere of radius $R$ is $ρ (r) = 20 \frac{r^{2}}{R^{2}}$ 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{π}{5}

2

3 π

3

\frac{3 π}{2}

4

π

Explanation:

D Given,
Density of a solid sphere $[ρ (r)] = 20 \frac{r^{2}}{R^{2}}$
Distance from centre $= r$
We know,
Volume of a spherical shell of radius $r$ and thickness $dr$ is,
$dV = 4 π r^{2} dr$
And, Density of a solid sphere $[ρ (r)] = \frac{dM}{dV}$
Then,
$dM = \frac{20 r^{2}}{R^{2}} (4 π r^{2} dr)$
$dM = \frac{80 π r^{4}}{R^{2}} dr$
$Mass of the shell (dM) = ρ . d V$
Total mass of complete solid sphere, $M = \int_{0}^{R} d M$
$M = \frac{80 π}{R^{2}} \int_{0}^{R} r^{4} dr$
$M = \frac{80 π}{R^{2}} {[\frac{r^{5}}{5}]}_{0}^{R}$
$M = \frac{80 π}{R^{2}} \times \frac{1}{5} [R^{5} - (0)^{5}]$
$M = \frac{16 π}{R^{2}} R^{5}$
$∴ M = 16 π R^{3}$
Gravitational field intensity at a distance $4 R$ from its centre $(E) = \frac{G M}{(4 R)^{2}}$
Then,
$E = \frac{G (16 π R^{3})}{16 R^{2}}$
$\Rightarrow E = G π R$
$∴ \frac{E}{GR} = π$

TS- EAMCET-04.05.2018

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}}{(2 g R - v_{0}^{2})}

2

\frac{{Rv}_{0}^{2}}{2 gR}

3

R

4

\frac{{Rv}_{0}^{2}}{(2 gR + v_{0}^{2})}

Explanation:

A Given,
Mass of body $= m$
Velocity of body $= v_{0}$
Mass of the earth $= M$
Radius of the earth $= R$
Free-fall acceleration on its surface $= g$ Height $= h$
We know,
Kinetic energy of the body (K.E. $)_{initial} = \frac{1}{2} {mv}_{0}^{2}$
And Potential energy of the body $(U)_{initial} = - \frac{G M m}{R}$ Now,
Potential energy of the body at height $h$ $(U)_{final} = - \frac{GMm}{(R + h)}$
According to law of conservation of energy,
$(K.E.)_{initial} + (U)_{initial} = (U)_{final}$
$\frac{1}{2} {mv}_{0}^{2} + (- \frac{GMm}{R}) = - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = \frac{GMm}{R} - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{1}{R} - \frac{1}{(R + h)})$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{R + h - R}{R (R + h)})$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R (R + h)}$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R^{2} (1 + \frac{h}{R})}$
Then,
$\frac{v_{0}^{2}}{2} = \frac{gh}{(1 + \frac{h}{R})} (∵ g = \frac{GM}{R^{2}})$
$\frac{v_{0}^{2}}{2} = \frac{ghR}{(R + h)}$
Now,
$v_{0}^{2} \times (R + h) = 2 g R h$
$v_{0}^{2} R + v_{0}^{2} h = 2 g R h$
$2 g R h - v_{0}^{2} h = R v_{0}^{2}$
$h (2 g R - v_{0}^{2}) = R v_{0}^{2}$
Hence, $h = \frac{{Rv}_{0}^{2}}{2 gR - v_{0}^{2}}$

MHT CET 2020

Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $ρ (r) = ρ_{0} {(\frac{r}{R})}^{β}$ , where $ρ_{0}$ and $β$ are constants and $r$ 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 $β$ is

1 2

2 2.5

3 3

4 4

Explanation:

C
Mass enclosed in a sphere of radius $r, M_{r} = \int ρ d V$
$= \int ρ_{0} {(\frac{r}{R})}^{β} 4 π r^{2} d r$
$= {[\frac{4 π ρ_{0}}{R^{β}} \times \frac{r^{β + 3}}{β + 3}]}_{0}^{r}$
When $radius (r) = \frac{R}{2}$ ,
Mass enclosed in a sphere $M_{1} = \frac{4 π ρ_{0}}{R^{β}} \frac{{(\frac{R}{2})}^{β + 3}}{(β + 3)}$
$= \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
Gravitational field intensity $E_{1} = \frac{{GM}_{1}}{{(\frac{R}{2})}^{2}}$
$= \frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
When radius $(r) = R$ ,
Mass enclosed in a sphere,
$M_{2} = \frac{4 π ρ_{0}}{R^{β}} \cdot \frac{R^{β + 3}}{β + 3} = \frac{4 π ρ_{0} R^{3}}{β + 3}$
Gravitational field intensity $(E_{2}) = \frac{{GM}_{2}}{(2 R)^{2}}$
$= \frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{β + 3}$
If $\frac{E_{2}}{E_{1}} = 4$ ,
$\frac{\frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)}}{\frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}} = 4$
Now,
$\frac{2^{β + 3}}{16} = 4$
$2^{β + 3} = 64$
$2^{β} \cdot 2^{3} = 2^{6}$
$2^{β} = 2^{3}$
$β = 3$

TS- EAMCET-05.05.2018

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 {Mv}^{2}

2

1 / 2 {Mv}^{2}

3

3 / 2 {Mv}^{2}

4

{Mv}^{2}

Explanation:

A Given,
Speed of satellite $= v$
Radius of circular orbit of satellite $= r$
Mass of satellite $= M$
Let Mass of earth $= M_{e}$
Radius of earth $= R_{e}$
We know,
Kinetic energy of Satellite (K.E.) $= \frac{1}{2} {Mv}^{2}$
And Orbital velocity of satellite $(v) = \sqrt{{gR}_{e}}$
Squaring both sides,
$v^{2} = {gR}_{e}$
Then,
$K.E. = \frac{1}{2} Mg R_{e}$
Potential energy of a satellite
$P.E. = - \frac{{GM}_{e} M}{R_{e}}$
In numerator and denominator multiply by $R_{e}$ ,
$P.E. = - \frac{{GM}_{e} M \cdot R_{e}}{R_{e}^{2}}$
Then,
$P.E. = - {gMR}_{e} (∵ g = \frac{{GM}_{e}}{R_{e}^{2}})$
Now,
Total Energy $=$ K.E. + P.E.
$= \frac{1}{2} {MgR}_{e} - {MgR}_{e}$
$= - \frac{1}{2} {MgR}_{e} [∵ v^{2} = {gR}_{e}]$
$= - \frac{1}{2} {Mv}^{2}$

MP PMT 2001

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 $g$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1

mgh

2

\frac{4}{5} mgh

3

\frac{5}{6} mgh

4

\frac{6}{7} mgh

Explanation:

C Given, Mass of earth $= M$
Mass of body $= m$
Radius of earth $= R$
We know,
Potential Energy on the earth's surface (P.E. $)_{1} = - \frac{G M m}{R}$
And Potential Energy at height $(h) = \frac{R}{5}$ from the earth's surface $(P.E.)_{2} = - \frac{GMm}{(R + \frac{R}{5})}$
$(P.E.)_{2} = - \frac{GMm}{(\frac{5 R + R}{5})}$
$(P.E.)_{2} = \frac{- 5 GMm}{6 R}$
Increase in potential energy (P.E. $) = (P.E.)_{2} - (P.E.)_{1}$
$P.E. = - \frac{5 GMm}{6 R} - (- \frac{GMm}{R})$
$P.E. = - \frac{5 GMm}{6 R} + \frac{GMm}{R}$
$P.E. = \frac{- 5 GMm + 6 GMm}{6 R}$
$P.E. = \frac{GMm}{6 R}$
Multiplying $R$ in numerator and denominator,
$P.E. = \frac{GMmR}{6 R^{2}}$
Then,
$\begin{array}{lr} P.E. = \frac{gmR}{6} & (∵ g = \frac{GM}{R^{2}}) \\ P.E. = \frac{gm (5 h)}{6} & [\begin{array}{l} ∵ h = \frac{R}{5} \\ R = 5 h \end{array}] \end{array}$
Hence, P.E. $= \frac{5}{6} mgh$

Manipal UGET -2020

Gravitation

138475 The density of a solid sphere of radius $R$ is $ρ (r) = 20 \frac{r^{2}}{R^{2}}$ 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{π}{5}

2

3 π

3

\frac{3 π}{2}

4

π

Explanation:

D Given,
Density of a solid sphere $[ρ (r)] = 20 \frac{r^{2}}{R^{2}}$
Distance from centre $= r$
We know,
Volume of a spherical shell of radius $r$ and thickness $dr$ is,
$dV = 4 π r^{2} dr$
And, Density of a solid sphere $[ρ (r)] = \frac{dM}{dV}$
Then,
$dM = \frac{20 r^{2}}{R^{2}} (4 π r^{2} dr)$
$dM = \frac{80 π r^{4}}{R^{2}} dr$
$Mass of the shell (dM) = ρ . d V$
Total mass of complete solid sphere, $M = \int_{0}^{R} d M$
$M = \frac{80 π}{R^{2}} \int_{0}^{R} r^{4} dr$
$M = \frac{80 π}{R^{2}} {[\frac{r^{5}}{5}]}_{0}^{R}$
$M = \frac{80 π}{R^{2}} \times \frac{1}{5} [R^{5} - (0)^{5}]$
$M = \frac{16 π}{R^{2}} R^{5}$
$∴ M = 16 π R^{3}$
Gravitational field intensity at a distance $4 R$ from its centre $(E) = \frac{G M}{(4 R)^{2}}$
Then,
$E = \frac{G (16 π R^{3})}{16 R^{2}}$
$\Rightarrow E = G π R$
$∴ \frac{E}{GR} = π$

TS- EAMCET-04.05.2018

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}}{(2 g R - v_{0}^{2})}

2

\frac{{Rv}_{0}^{2}}{2 gR}

3

R

4

\frac{{Rv}_{0}^{2}}{(2 gR + v_{0}^{2})}

Explanation:

A Given,
Mass of body $= m$
Velocity of body $= v_{0}$
Mass of the earth $= M$
Radius of the earth $= R$
Free-fall acceleration on its surface $= g$ Height $= h$
We know,
Kinetic energy of the body (K.E. $)_{initial} = \frac{1}{2} {mv}_{0}^{2}$
And Potential energy of the body $(U)_{initial} = - \frac{G M m}{R}$ Now,
Potential energy of the body at height $h$ $(U)_{final} = - \frac{GMm}{(R + h)}$
According to law of conservation of energy,
$(K.E.)_{initial} + (U)_{initial} = (U)_{final}$
$\frac{1}{2} {mv}_{0}^{2} + (- \frac{GMm}{R}) = - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = \frac{GMm}{R} - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{1}{R} - \frac{1}{(R + h)})$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{R + h - R}{R (R + h)})$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R (R + h)}$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R^{2} (1 + \frac{h}{R})}$
Then,
$\frac{v_{0}^{2}}{2} = \frac{gh}{(1 + \frac{h}{R})} (∵ g = \frac{GM}{R^{2}})$
$\frac{v_{0}^{2}}{2} = \frac{ghR}{(R + h)}$
Now,
$v_{0}^{2} \times (R + h) = 2 g R h$
$v_{0}^{2} R + v_{0}^{2} h = 2 g R h$
$2 g R h - v_{0}^{2} h = R v_{0}^{2}$
$h (2 g R - v_{0}^{2}) = R v_{0}^{2}$
Hence, $h = \frac{{Rv}_{0}^{2}}{2 gR - v_{0}^{2}}$

MHT CET 2020

Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $ρ (r) = ρ_{0} {(\frac{r}{R})}^{β}$ , where $ρ_{0}$ and $β$ are constants and $r$ 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 $β$ is

1 2

2 2.5

3 3

4 4

Explanation:

C
Mass enclosed in a sphere of radius $r, M_{r} = \int ρ d V$
$= \int ρ_{0} {(\frac{r}{R})}^{β} 4 π r^{2} d r$
$= {[\frac{4 π ρ_{0}}{R^{β}} \times \frac{r^{β + 3}}{β + 3}]}_{0}^{r}$
When $radius (r) = \frac{R}{2}$ ,
Mass enclosed in a sphere $M_{1} = \frac{4 π ρ_{0}}{R^{β}} \frac{{(\frac{R}{2})}^{β + 3}}{(β + 3)}$
$= \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
Gravitational field intensity $E_{1} = \frac{{GM}_{1}}{{(\frac{R}{2})}^{2}}$
$= \frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
When radius $(r) = R$ ,
Mass enclosed in a sphere,
$M_{2} = \frac{4 π ρ_{0}}{R^{β}} \cdot \frac{R^{β + 3}}{β + 3} = \frac{4 π ρ_{0} R^{3}}{β + 3}$
Gravitational field intensity $(E_{2}) = \frac{{GM}_{2}}{(2 R)^{2}}$
$= \frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{β + 3}$
If $\frac{E_{2}}{E_{1}} = 4$ ,
$\frac{\frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)}}{\frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}} = 4$
Now,
$\frac{2^{β + 3}}{16} = 4$
$2^{β + 3} = 64$
$2^{β} \cdot 2^{3} = 2^{6}$
$2^{β} = 2^{3}$
$β = 3$

TS- EAMCET-05.05.2018

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 {Mv}^{2}

2

1 / 2 {Mv}^{2}

3

3 / 2 {Mv}^{2}

4

{Mv}^{2}

Explanation:

A Given,
Speed of satellite $= v$
Radius of circular orbit of satellite $= r$
Mass of satellite $= M$
Let Mass of earth $= M_{e}$
Radius of earth $= R_{e}$
We know,
Kinetic energy of Satellite (K.E.) $= \frac{1}{2} {Mv}^{2}$
And Orbital velocity of satellite $(v) = \sqrt{{gR}_{e}}$
Squaring both sides,
$v^{2} = {gR}_{e}$
Then,
$K.E. = \frac{1}{2} Mg R_{e}$
Potential energy of a satellite
$P.E. = - \frac{{GM}_{e} M}{R_{e}}$
In numerator and denominator multiply by $R_{e}$ ,
$P.E. = - \frac{{GM}_{e} M \cdot R_{e}}{R_{e}^{2}}$
Then,
$P.E. = - {gMR}_{e} (∵ g = \frac{{GM}_{e}}{R_{e}^{2}})$
Now,
Total Energy $=$ K.E. + P.E.
$= \frac{1}{2} {MgR}_{e} - {MgR}_{e}$
$= - \frac{1}{2} {MgR}_{e} [∵ v^{2} = {gR}_{e}]$
$= - \frac{1}{2} {Mv}^{2}$

MP PMT 2001

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 $g$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1

mgh

2

\frac{4}{5} mgh

3

\frac{5}{6} mgh

4

\frac{6}{7} mgh

Explanation:

C Given, Mass of earth $= M$
Mass of body $= m$
Radius of earth $= R$
We know,
Potential Energy on the earth's surface (P.E. $)_{1} = - \frac{G M m}{R}$
And Potential Energy at height $(h) = \frac{R}{5}$ from the earth's surface $(P.E.)_{2} = - \frac{GMm}{(R + \frac{R}{5})}$
$(P.E.)_{2} = - \frac{GMm}{(\frac{5 R + R}{5})}$
$(P.E.)_{2} = \frac{- 5 GMm}{6 R}$
Increase in potential energy (P.E. $) = (P.E.)_{2} - (P.E.)_{1}$
$P.E. = - \frac{5 GMm}{6 R} - (- \frac{GMm}{R})$
$P.E. = - \frac{5 GMm}{6 R} + \frac{GMm}{R}$
$P.E. = \frac{- 5 GMm + 6 GMm}{6 R}$
$P.E. = \frac{GMm}{6 R}$
Multiplying $R$ in numerator and denominator,
$P.E. = \frac{GMmR}{6 R^{2}}$
Then,
$\begin{array}{lr} P.E. = \frac{gmR}{6} & (∵ g = \frac{GM}{R^{2}}) \\ P.E. = \frac{gm (5 h)}{6} & [\begin{array}{l} ∵ h = \frac{R}{5} \\ R = 5 h \end{array}] \end{array}$
Hence, P.E. $= \frac{5}{6} mgh$

Manipal UGET -2020

Gravitation

138475 The density of a solid sphere of radius $R$ is $ρ (r) = 20 \frac{r^{2}}{R^{2}}$ 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{π}{5}

2

3 π

3

\frac{3 π}{2}

4

π

Explanation:

D Given,
Density of a solid sphere $[ρ (r)] = 20 \frac{r^{2}}{R^{2}}$
Distance from centre $= r$
We know,
Volume of a spherical shell of radius $r$ and thickness $dr$ is,
$dV = 4 π r^{2} dr$
And, Density of a solid sphere $[ρ (r)] = \frac{dM}{dV}$
Then,
$dM = \frac{20 r^{2}}{R^{2}} (4 π r^{2} dr)$
$dM = \frac{80 π r^{4}}{R^{2}} dr$
$Mass of the shell (dM) = ρ . d V$
Total mass of complete solid sphere, $M = \int_{0}^{R} d M$
$M = \frac{80 π}{R^{2}} \int_{0}^{R} r^{4} dr$
$M = \frac{80 π}{R^{2}} {[\frac{r^{5}}{5}]}_{0}^{R}$
$M = \frac{80 π}{R^{2}} \times \frac{1}{5} [R^{5} - (0)^{5}]$
$M = \frac{16 π}{R^{2}} R^{5}$
$∴ M = 16 π R^{3}$
Gravitational field intensity at a distance $4 R$ from its centre $(E) = \frac{G M}{(4 R)^{2}}$
Then,
$E = \frac{G (16 π R^{3})}{16 R^{2}}$
$\Rightarrow E = G π R$
$∴ \frac{E}{GR} = π$

TS- EAMCET-04.05.2018

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}}{(2 g R - v_{0}^{2})}

2

\frac{{Rv}_{0}^{2}}{2 gR}

3

R

4

\frac{{Rv}_{0}^{2}}{(2 gR + v_{0}^{2})}

Explanation:

A Given,
Mass of body $= m$
Velocity of body $= v_{0}$
Mass of the earth $= M$
Radius of the earth $= R$
Free-fall acceleration on its surface $= g$ Height $= h$
We know,
Kinetic energy of the body (K.E. $)_{initial} = \frac{1}{2} {mv}_{0}^{2}$
And Potential energy of the body $(U)_{initial} = - \frac{G M m}{R}$ Now,
Potential energy of the body at height $h$ $(U)_{final} = - \frac{GMm}{(R + h)}$
According to law of conservation of energy,
$(K.E.)_{initial} + (U)_{initial} = (U)_{final}$
$\frac{1}{2} {mv}_{0}^{2} + (- \frac{GMm}{R}) = - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = \frac{GMm}{R} - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{1}{R} - \frac{1}{(R + h)})$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{R + h - R}{R (R + h)})$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R (R + h)}$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R^{2} (1 + \frac{h}{R})}$
Then,
$\frac{v_{0}^{2}}{2} = \frac{gh}{(1 + \frac{h}{R})} (∵ g = \frac{GM}{R^{2}})$
$\frac{v_{0}^{2}}{2} = \frac{ghR}{(R + h)}$
Now,
$v_{0}^{2} \times (R + h) = 2 g R h$
$v_{0}^{2} R + v_{0}^{2} h = 2 g R h$
$2 g R h - v_{0}^{2} h = R v_{0}^{2}$
$h (2 g R - v_{0}^{2}) = R v_{0}^{2}$
Hence, $h = \frac{{Rv}_{0}^{2}}{2 gR - v_{0}^{2}}$

MHT CET 2020

Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $ρ (r) = ρ_{0} {(\frac{r}{R})}^{β}$ , where $ρ_{0}$ and $β$ are constants and $r$ 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 $β$ is

1 2

2 2.5

3 3

4 4

Explanation:

C
Mass enclosed in a sphere of radius $r, M_{r} = \int ρ d V$
$= \int ρ_{0} {(\frac{r}{R})}^{β} 4 π r^{2} d r$
$= {[\frac{4 π ρ_{0}}{R^{β}} \times \frac{r^{β + 3}}{β + 3}]}_{0}^{r}$
When $radius (r) = \frac{R}{2}$ ,
Mass enclosed in a sphere $M_{1} = \frac{4 π ρ_{0}}{R^{β}} \frac{{(\frac{R}{2})}^{β + 3}}{(β + 3)}$
$= \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
Gravitational field intensity $E_{1} = \frac{{GM}_{1}}{{(\frac{R}{2})}^{2}}$
$= \frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
When radius $(r) = R$ ,
Mass enclosed in a sphere,
$M_{2} = \frac{4 π ρ_{0}}{R^{β}} \cdot \frac{R^{β + 3}}{β + 3} = \frac{4 π ρ_{0} R^{3}}{β + 3}$
Gravitational field intensity $(E_{2}) = \frac{{GM}_{2}}{(2 R)^{2}}$
$= \frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{β + 3}$
If $\frac{E_{2}}{E_{1}} = 4$ ,
$\frac{\frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)}}{\frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}} = 4$
Now,
$\frac{2^{β + 3}}{16} = 4$
$2^{β + 3} = 64$
$2^{β} \cdot 2^{3} = 2^{6}$
$2^{β} = 2^{3}$
$β = 3$

TS- EAMCET-05.05.2018

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 {Mv}^{2}

2

1 / 2 {Mv}^{2}

3

3 / 2 {Mv}^{2}

4

{Mv}^{2}

Explanation:

A Given,
Speed of satellite $= v$
Radius of circular orbit of satellite $= r$
Mass of satellite $= M$
Let Mass of earth $= M_{e}$
Radius of earth $= R_{e}$
We know,
Kinetic energy of Satellite (K.E.) $= \frac{1}{2} {Mv}^{2}$
And Orbital velocity of satellite $(v) = \sqrt{{gR}_{e}}$
Squaring both sides,
$v^{2} = {gR}_{e}$
Then,
$K.E. = \frac{1}{2} Mg R_{e}$
Potential energy of a satellite
$P.E. = - \frac{{GM}_{e} M}{R_{e}}$
In numerator and denominator multiply by $R_{e}$ ,
$P.E. = - \frac{{GM}_{e} M \cdot R_{e}}{R_{e}^{2}}$
Then,
$P.E. = - {gMR}_{e} (∵ g = \frac{{GM}_{e}}{R_{e}^{2}})$
Now,
Total Energy $=$ K.E. + P.E.
$= \frac{1}{2} {MgR}_{e} - {MgR}_{e}$
$= - \frac{1}{2} {MgR}_{e} [∵ v^{2} = {gR}_{e}]$
$= - \frac{1}{2} {Mv}^{2}$

MP PMT 2001

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 $g$ is acceleration due to gravity at the earth's surface, the increase in potential energy will be

1

mgh

2

\frac{4}{5} mgh

3

\frac{5}{6} mgh

4

\frac{6}{7} mgh

Explanation:

C Given, Mass of earth $= M$
Mass of body $= m$
Radius of earth $= R$
We know,
Potential Energy on the earth's surface (P.E. $)_{1} = - \frac{G M m}{R}$
And Potential Energy at height $(h) = \frac{R}{5}$ from the earth's surface $(P.E.)_{2} = - \frac{GMm}{(R + \frac{R}{5})}$
$(P.E.)_{2} = - \frac{GMm}{(\frac{5 R + R}{5})}$
$(P.E.)_{2} = \frac{- 5 GMm}{6 R}$
Increase in potential energy (P.E. $) = (P.E.)_{2} - (P.E.)_{1}$
$P.E. = - \frac{5 GMm}{6 R} - (- \frac{GMm}{R})$
$P.E. = - \frac{5 GMm}{6 R} + \frac{GMm}{R}$
$P.E. = \frac{- 5 GMm + 6 GMm}{6 R}$
$P.E. = \frac{GMm}{6 R}$
Multiplying $R$ in numerator and denominator,
$P.E. = \frac{GMmR}{6 R^{2}}$
Then,
$\begin{array}{lr} P.E. = \frac{gmR}{6} & (∵ g = \frac{GM}{R^{2}}) \\ P.E. = \frac{gm (5 h)}{6} & [\begin{array}{l} ∵ h = \frac{R}{5} \\ R = 5 h \end{array}] \end{array}$
Hence, P.E. $= \frac{5}{6} mgh$

Manipal UGET -2020

Gravitation

138475 The density of a solid sphere of radius $R$ is $ρ (r) = 20 \frac{r^{2}}{R^{2}}$ 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{π}{5}

2

3 π

3

\frac{3 π}{2}

4

π

Explanation:

D Given,
Density of a solid sphere $[ρ (r)] = 20 \frac{r^{2}}{R^{2}}$
Distance from centre $= r$
We know,
Volume of a spherical shell of radius $r$ and thickness $dr$ is,
$dV = 4 π r^{2} dr$
And, Density of a solid sphere $[ρ (r)] = \frac{dM}{dV}$
Then,
$dM = \frac{20 r^{2}}{R^{2}} (4 π r^{2} dr)$
$dM = \frac{80 π r^{4}}{R^{2}} dr$
$Mass of the shell (dM) = ρ . d V$
Total mass of complete solid sphere, $M = \int_{0}^{R} d M$
$M = \frac{80 π}{R^{2}} \int_{0}^{R} r^{4} dr$
$M = \frac{80 π}{R^{2}} {[\frac{r^{5}}{5}]}_{0}^{R}$
$M = \frac{80 π}{R^{2}} \times \frac{1}{5} [R^{5} - (0)^{5}]$
$M = \frac{16 π}{R^{2}} R^{5}$
$∴ M = 16 π R^{3}$
Gravitational field intensity at a distance $4 R$ from its centre $(E) = \frac{G M}{(4 R)^{2}}$
Then,
$E = \frac{G (16 π R^{3})}{16 R^{2}}$
$\Rightarrow E = G π R$
$∴ \frac{E}{GR} = π$

TS- EAMCET-04.05.2018

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}}{(2 g R - v_{0}^{2})}

2

\frac{{Rv}_{0}^{2}}{2 gR}

3

R

4

\frac{{Rv}_{0}^{2}}{(2 gR + v_{0}^{2})}

Explanation:

A Given,
Mass of body $= m$
Velocity of body $= v_{0}$
Mass of the earth $= M$
Radius of the earth $= R$
Free-fall acceleration on its surface $= g$ Height $= h$
We know,
Kinetic energy of the body (K.E. $)_{initial} = \frac{1}{2} {mv}_{0}^{2}$
And Potential energy of the body $(U)_{initial} = - \frac{G M m}{R}$ Now,
Potential energy of the body at height $h$ $(U)_{final} = - \frac{GMm}{(R + h)}$
According to law of conservation of energy,
$(K.E.)_{initial} + (U)_{initial} = (U)_{final}$
$\frac{1}{2} {mv}_{0}^{2} + (- \frac{GMm}{R}) = - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = \frac{GMm}{R} - \frac{GMm}{(R + h)}$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{1}{R} - \frac{1}{(R + h)})$
$\frac{1}{2} {mv}_{0}^{2} = GMm (\frac{R + h - R}{R (R + h)})$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R (R + h)}$
$\frac{v_{0}^{2}}{2} = \frac{GMh}{R^{2} (1 + \frac{h}{R})}$
Then,
$\frac{v_{0}^{2}}{2} = \frac{gh}{(1 + \frac{h}{R})} (∵ g = \frac{GM}{R^{2}})$
$\frac{v_{0}^{2}}{2} = \frac{ghR}{(R + h)}$
Now,
$v_{0}^{2} \times (R + h) = 2 g R h$
$v_{0}^{2} R + v_{0}^{2} h = 2 g R h$
$2 g R h - v_{0}^{2} h = R v_{0}^{2}$
$h (2 g R - v_{0}^{2}) = R v_{0}^{2}$
Hence, $h = \frac{{Rv}_{0}^{2}}{2 gR - v_{0}^{2}}$

MHT CET 2020

Gravitation

138477 The mass density inside a solid sphere of radius $r$ varies as $ρ (r) = ρ_{0} {(\frac{r}{R})}^{β}$ , where $ρ_{0}$ and $β$ are constants and $r$ 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 $β$ is

1 2

2 2.5

3 3

4 4

Explanation:

C
Mass enclosed in a sphere of radius $r, M_{r} = \int ρ d V$
$= \int ρ_{0} {(\frac{r}{R})}^{β} 4 π r^{2} d r$
$= {[\frac{4 π ρ_{0}}{R^{β}} \times \frac{r^{β + 3}}{β + 3}]}_{0}^{r}$
When $radius (r) = \frac{R}{2}$ ,
Mass enclosed in a sphere $M_{1} = \frac{4 π ρ_{0}}{R^{β}} \frac{{(\frac{R}{2})}^{β + 3}}{(β + 3)}$
$= \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
Gravitational field intensity $E_{1} = \frac{{GM}_{1}}{{(\frac{R}{2})}^{2}}$
$= \frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}$
When radius $(r) = R$ ,
Mass enclosed in a sphere,
$M_{2} = \frac{4 π ρ_{0}}{R^{β}} \cdot \frac{R^{β + 3}}{β + 3} = \frac{4 π ρ_{0} R^{3}}{β + 3}$
Gravitational field intensity $(E_{2}) = \frac{{GM}_{2}}{(2 R)^{2}}$
$= \frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{β + 3}$
If $\frac{E_{2}}{E_{1}} = 4$ ,
$\frac{\frac{G}{4 R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)}}{\frac{4 G}{R^{2}} \times \frac{4 π ρ_{0} R^{3}}{(β + 3)} \times \frac{1}{2^{β + 3}}} = 4$
Now,
$\frac{2^{β + 3}}{16} = 4$
$2^{β + 3} = 64$
$2^{β} \cdot 2^{3} = 2^{6}$
$2^{β} = 2^{3}$
$β = 3$

TS- EAMCET-05.05.2018

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: