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Gravitation

138483 The gravitational field due to a mass distribution is $E = \frac{k}{x^{3}}$ in the $x$ -direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is-
[Taking gravitational potential to be zero at infinity]

1

\frac{k}{x}

2

\frac{k}{x^{3}}

3

\frac{k}{2 x^{2}}

4

\frac{k}{3 x^{3}}

Explanation:

C According to question,
The gravitational field due to a mass distribution in $x$ direction is-
$E = \frac{k}{x^{3}}$
Where, $k$ is a constant
Gravitational force $= \int_{x}^{\infty} E d x$
Putting the value of $E$ from equation (i), we get
$= \int_{x}^{\infty} \frac{k}{x^{3}} dx$
$= k \int_{x}^{\infty} x^{- 3} dx$
$= k {[\frac{x^{- 3 + 1}}{- 3 + 1}]}_{x}^{\infty}$
$= k {[\frac{x^{- 2}}{- 2}]}_{x}^{\infty}$
$= - \frac{k}{2} {[x^{- 2}]}_{x}^{\infty}$
$= \frac{k}{2 x^{2}}$
Hence, the value of gravitational potential at a distance $x$ is $k / 2 x^{2}$ .

BCECE-2014

Gravitation

138484 The gravitational potential at a place varies inversely with $x^{2} ($ i.e. $V = k / x^{2})$ , the gravitational field at that place is

1

2 k / x^{3}

2

- 2 k / x^{3}

3

k / x

4

- k / x

Explanation:

A Given that,
Gravitational potential-
$V = \frac{k}{x^{2}}$
Gravitational field-
$I = \frac{- d}{dx} (V)$
Putting the value of $V$ from equation (i), we get
$I = \frac{- d}{d x} (\frac{k}{x^{2}})$ $= \frac{- d}{dx} ({kx}^{- 2})$
$= - k \times \frac{- 2}{x^{3}}$
$I = \frac{2 k}{x^{3}}$

VITEEE-2013

Gravitation

138485 A projectile is fired from the earth vertically with a velocity $kv$ where $v$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is :
Where $k < 1$ .

1

\frac{R}{k^{2}}

2

\frac{R}{1 + k^{2}}

3

\frac{R}{1 - k^{2}}

4

\frac{kR}{1 + k^{2}}

Explanation:

C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $h$ , then according to the law of conservation of energy,
$\frac{1}{2} {mv}^{2} = \frac{mgh}{1 + \frac{h}{R}}$
Here, $v = {kv}_{e}$
$v = k \sqrt{2 gR}$
$(∵ v_{e} = \sqrt{2 gR})$
Squaring both side, $v^{2} = 2 k^{2} gR$
Putting the value of $v^{2}$ in equation (i), we get
$\frac{1}{2} {mk}^{2} 2 gR = \frac{mg (r - R)}{[1 + \frac{(r - R)}{R}]} {h = r - R}$
$k^{2} R = \frac{(r - R)}{[1 + \frac{(r - R)}{R}]}$
$k^{2} R [1 + \frac{r - R}{R}] = r - R$
or $k^{2} r = r - R$
$R = r - k^{2} r$
$R = r (1 - k^{2})$
$r = \frac{R}{1 - k^{2}}$

SCRA-2012

Gravitation

138486 A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :

1

\frac{mgR}{2}

2

mgR

3

2 mgR

4

\frac{mgR}{4}

Explanation:

A At Earth surface gravitational potential energy of the mass
$U_{E} = \frac{- GMm}{R}$
At target gravitational force-
$U_{T} = \frac{- GMm}{h + R} (∵ h = R)$
$U_{T} = \frac{- GMm}{R + R}$
$U_{T} = \frac{- GMm}{2 R}$
From equation (i) and (ii), we get
Work done $(W) = \frac{- GMm}{2 R} - (\frac{- GMm}{R})$
$= \frac{- GMm + 2 GMm}{2 R} = \frac{GMm}{2 R} (∵ g = \frac{GM}{R^{2}} \Rightarrow \frac{GM}{R} = gR)$
$∴ W = \frac{mgR}{2}$

Karnataka CET-2018

Gravitation

138483 The gravitational field due to a mass distribution is $E = \frac{k}{x^{3}}$ in the $x$ -direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is-
[Taking gravitational potential to be zero at infinity]

1

\frac{k}{x}

2

\frac{k}{x^{3}}

3

\frac{k}{2 x^{2}}

4

\frac{k}{3 x^{3}}

Explanation:

C According to question,
The gravitational field due to a mass distribution in $x$ direction is-
$E = \frac{k}{x^{3}}$
Where, $k$ is a constant
Gravitational force $= \int_{x}^{\infty} E d x$
Putting the value of $E$ from equation (i), we get
$= \int_{x}^{\infty} \frac{k}{x^{3}} dx$
$= k \int_{x}^{\infty} x^{- 3} dx$
$= k {[\frac{x^{- 3 + 1}}{- 3 + 1}]}_{x}^{\infty}$
$= k {[\frac{x^{- 2}}{- 2}]}_{x}^{\infty}$
$= - \frac{k}{2} {[x^{- 2}]}_{x}^{\infty}$
$= \frac{k}{2 x^{2}}$
Hence, the value of gravitational potential at a distance $x$ is $k / 2 x^{2}$ .

BCECE-2014

Gravitation

138484 The gravitational potential at a place varies inversely with $x^{2} ($ i.e. $V = k / x^{2})$ , the gravitational field at that place is

1

2 k / x^{3}

2

- 2 k / x^{3}

3

k / x

4

- k / x

Explanation:

A Given that,
Gravitational potential-
$V = \frac{k}{x^{2}}$
Gravitational field-
$I = \frac{- d}{dx} (V)$
Putting the value of $V$ from equation (i), we get
$I = \frac{- d}{d x} (\frac{k}{x^{2}})$ $= \frac{- d}{dx} ({kx}^{- 2})$
$= - k \times \frac{- 2}{x^{3}}$
$I = \frac{2 k}{x^{3}}$

VITEEE-2013

Gravitation

138485 A projectile is fired from the earth vertically with a velocity $kv$ where $v$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is :
Where $k < 1$ .

1

\frac{R}{k^{2}}

2

\frac{R}{1 + k^{2}}

3

\frac{R}{1 - k^{2}}

4

\frac{kR}{1 + k^{2}}

Explanation:

C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $h$ , then according to the law of conservation of energy,
$\frac{1}{2} {mv}^{2} = \frac{mgh}{1 + \frac{h}{R}}$
Here, $v = {kv}_{e}$
$v = k \sqrt{2 gR}$
$(∵ v_{e} = \sqrt{2 gR})$
Squaring both side, $v^{2} = 2 k^{2} gR$
Putting the value of $v^{2}$ in equation (i), we get
$\frac{1}{2} {mk}^{2} 2 gR = \frac{mg (r - R)}{[1 + \frac{(r - R)}{R}]} {h = r - R}$
$k^{2} R = \frac{(r - R)}{[1 + \frac{(r - R)}{R}]}$
$k^{2} R [1 + \frac{r - R}{R}] = r - R$
or $k^{2} r = r - R$
$R = r - k^{2} r$
$R = r (1 - k^{2})$
$r = \frac{R}{1 - k^{2}}$

SCRA-2012

Gravitation

138486 A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :

1

\frac{mgR}{2}

2

mgR

3

2 mgR

4

\frac{mgR}{4}

Explanation:

A At Earth surface gravitational potential energy of the mass
$U_{E} = \frac{- GMm}{R}$
At target gravitational force-
$U_{T} = \frac{- GMm}{h + R} (∵ h = R)$
$U_{T} = \frac{- GMm}{R + R}$
$U_{T} = \frac{- GMm}{2 R}$
From equation (i) and (ii), we get
Work done $(W) = \frac{- GMm}{2 R} - (\frac{- GMm}{R})$
$= \frac{- GMm + 2 GMm}{2 R} = \frac{GMm}{2 R} (∵ g = \frac{GM}{R^{2}} \Rightarrow \frac{GM}{R} = gR)$
$∴ W = \frac{mgR}{2}$

Karnataka CET-2018

Gravitation

138483 The gravitational field due to a mass distribution is $E = \frac{k}{x^{3}}$ in the $x$ -direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is-
[Taking gravitational potential to be zero at infinity]

1

\frac{k}{x}

2

\frac{k}{x^{3}}

3

\frac{k}{2 x^{2}}

4

\frac{k}{3 x^{3}}

Explanation:

C According to question,
The gravitational field due to a mass distribution in $x$ direction is-
$E = \frac{k}{x^{3}}$
Where, $k$ is a constant
Gravitational force $= \int_{x}^{\infty} E d x$
Putting the value of $E$ from equation (i), we get
$= \int_{x}^{\infty} \frac{k}{x^{3}} dx$
$= k \int_{x}^{\infty} x^{- 3} dx$
$= k {[\frac{x^{- 3 + 1}}{- 3 + 1}]}_{x}^{\infty}$
$= k {[\frac{x^{- 2}}{- 2}]}_{x}^{\infty}$
$= - \frac{k}{2} {[x^{- 2}]}_{x}^{\infty}$
$= \frac{k}{2 x^{2}}$
Hence, the value of gravitational potential at a distance $x$ is $k / 2 x^{2}$ .

BCECE-2014

Gravitation

138484 The gravitational potential at a place varies inversely with $x^{2} ($ i.e. $V = k / x^{2})$ , the gravitational field at that place is

1

2 k / x^{3}

2

- 2 k / x^{3}

3

k / x

4

- k / x

Explanation:

A Given that,
Gravitational potential-
$V = \frac{k}{x^{2}}$
Gravitational field-
$I = \frac{- d}{dx} (V)$
Putting the value of $V$ from equation (i), we get
$I = \frac{- d}{d x} (\frac{k}{x^{2}})$ $= \frac{- d}{dx} ({kx}^{- 2})$
$= - k \times \frac{- 2}{x^{3}}$
$I = \frac{2 k}{x^{3}}$

VITEEE-2013

Gravitation

138485 A projectile is fired from the earth vertically with a velocity $kv$ where $v$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is :
Where $k < 1$ .

1

\frac{R}{k^{2}}

2

\frac{R}{1 + k^{2}}

3

\frac{R}{1 - k^{2}}

4

\frac{kR}{1 + k^{2}}

Explanation:

C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $h$ , then according to the law of conservation of energy,
$\frac{1}{2} {mv}^{2} = \frac{mgh}{1 + \frac{h}{R}}$
Here, $v = {kv}_{e}$
$v = k \sqrt{2 gR}$
$(∵ v_{e} = \sqrt{2 gR})$
Squaring both side, $v^{2} = 2 k^{2} gR$
Putting the value of $v^{2}$ in equation (i), we get
$\frac{1}{2} {mk}^{2} 2 gR = \frac{mg (r - R)}{[1 + \frac{(r - R)}{R}]} {h = r - R}$
$k^{2} R = \frac{(r - R)}{[1 + \frac{(r - R)}{R}]}$
$k^{2} R [1 + \frac{r - R}{R}] = r - R$
or $k^{2} r = r - R$
$R = r - k^{2} r$
$R = r (1 - k^{2})$
$r = \frac{R}{1 - k^{2}}$

SCRA-2012

Gravitation

138486 A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :

1

\frac{mgR}{2}

2

mgR

3

2 mgR

4

\frac{mgR}{4}

Explanation:

A At Earth surface gravitational potential energy of the mass
$U_{E} = \frac{- GMm}{R}$
At target gravitational force-
$U_{T} = \frac{- GMm}{h + R} (∵ h = R)$
$U_{T} = \frac{- GMm}{R + R}$
$U_{T} = \frac{- GMm}{2 R}$
From equation (i) and (ii), we get
Work done $(W) = \frac{- GMm}{2 R} - (\frac{- GMm}{R})$
$= \frac{- GMm + 2 GMm}{2 R} = \frac{GMm}{2 R} (∵ g = \frac{GM}{R^{2}} \Rightarrow \frac{GM}{R} = gR)$
$∴ W = \frac{mgR}{2}$

Karnataka CET-2018

Gravitation

138483 The gravitational field due to a mass distribution is $E = \frac{k}{x^{3}}$ in the $x$ -direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is-
[Taking gravitational potential to be zero at infinity]

1

\frac{k}{x}

2

\frac{k}{x^{3}}

3

\frac{k}{2 x^{2}}

4

\frac{k}{3 x^{3}}

Explanation:

C According to question,
The gravitational field due to a mass distribution in $x$ direction is-
$E = \frac{k}{x^{3}}$
Where, $k$ is a constant
Gravitational force $= \int_{x}^{\infty} E d x$
Putting the value of $E$ from equation (i), we get
$= \int_{x}^{\infty} \frac{k}{x^{3}} dx$
$= k \int_{x}^{\infty} x^{- 3} dx$
$= k {[\frac{x^{- 3 + 1}}{- 3 + 1}]}_{x}^{\infty}$
$= k {[\frac{x^{- 2}}{- 2}]}_{x}^{\infty}$
$= - \frac{k}{2} {[x^{- 2}]}_{x}^{\infty}$
$= \frac{k}{2 x^{2}}$
Hence, the value of gravitational potential at a distance $x$ is $k / 2 x^{2}$ .

BCECE-2014

Gravitation

138484 The gravitational potential at a place varies inversely with $x^{2} ($ i.e. $V = k / x^{2})$ , the gravitational field at that place is

1

2 k / x^{3}

2

- 2 k / x^{3}

3

k / x

4

- k / x

Explanation:

A Given that,
Gravitational potential-
$V = \frac{k}{x^{2}}$
Gravitational field-
$I = \frac{- d}{dx} (V)$
Putting the value of $V$ from equation (i), we get
$I = \frac{- d}{d x} (\frac{k}{x^{2}})$ $= \frac{- d}{dx} ({kx}^{- 2})$
$= - k \times \frac{- 2}{x^{3}}$
$I = \frac{2 k}{x^{3}}$

VITEEE-2013

Gravitation

138485 A projectile is fired from the earth vertically with a velocity $kv$ where $v$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is :
Where $k < 1$ .

1

\frac{R}{k^{2}}

2

\frac{R}{1 + k^{2}}

3

\frac{R}{1 - k^{2}}

4

\frac{kR}{1 + k^{2}}

Explanation:

C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $h$ , then according to the law of conservation of energy,
$\frac{1}{2} {mv}^{2} = \frac{mgh}{1 + \frac{h}{R}}$
Here, $v = {kv}_{e}$
$v = k \sqrt{2 gR}$
$(∵ v_{e} = \sqrt{2 gR})$
Squaring both side, $v^{2} = 2 k^{2} gR$
Putting the value of $v^{2}$ in equation (i), we get
$\frac{1}{2} {mk}^{2} 2 gR = \frac{mg (r - R)}{[1 + \frac{(r - R)}{R}]} {h = r - R}$
$k^{2} R = \frac{(r - R)}{[1 + \frac{(r - R)}{R}]}$
$k^{2} R [1 + \frac{r - R}{R}] = r - R$
or $k^{2} r = r - R$
$R = r - k^{2} r$
$R = r (1 - k^{2})$
$r = \frac{R}{1 - k^{2}}$

SCRA-2012

Gravitation

138486 A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :

1

\frac{mgR}{2}

2

mgR

3

2 mgR

4

\frac{mgR}{4}

Explanation:

A At Earth surface gravitational potential energy of the mass
$U_{E} = \frac{- GMm}{R}$
At target gravitational force-
$U_{T} = \frac{- GMm}{h + R} (∵ h = R)$
$U_{T} = \frac{- GMm}{R + R}$
$U_{T} = \frac{- GMm}{2 R}$
From equation (i) and (ii), we get
Work done $(W) = \frac{- GMm}{2 R} - (\frac{- GMm}{R})$
$= \frac{- GMm + 2 GMm}{2 R} = \frac{GMm}{2 R} (∵ g = \frac{GM}{R^{2}} \Rightarrow \frac{GM}{R} = gR)$
$∴ W = \frac{mgR}{2}$

Karnataka CET-2018