QBManager

Gravitation

138376 What will be the acceleration due to gravity at a depth $d$ , where $g$ is acceleration due to gravity on the surface of earth ?

1

\frac{g}{{[1 + \frac{d}{R}]}^{2}}

2

g [1 - \frac{2 d}{R}]

3

\frac{g}{{[1 - \frac{d}{R}]}^{2}}

4

g [1 - \frac{d}{R}]

Explanation:

D Variation of $g$ with depth

In this figure
$m = mass$
$d = distance below the earth's surface$
We know that on the surface of the earth
$g = \frac{4}{3} π ρ RG [∴ g = \frac{GM}{R^{2}}]$
At a distance (d) below the earth's surface
$g_{d} = \frac{4}{3} π ρ (R - d) G$
$\frac{g}{g_{d}} = \frac{R}{(R - d)}$
$g_{d} = (1 - \frac{d}{R}) g$
When, $d = 0$ then value of $g$ on the surface
And
$g_{d} = g$
When, $d = R$ then value of $g$ at the centre of the earth.
$g_{d} = 0$

UP CPMT-2010

Gravitation

138377 The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be

1 cannot be compared

2 equal to

T

3 less than

T

4 more than

T

Explanation:

D Time Period $(T) = 2 π \sqrt{\frac{l}{g}}$
$T \propto \frac{1}{\sqrt{g}}$
Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $R$ ,
Then,
$g^{'} = g (1 - \frac{h}{R})$
Hence, $g$ decreases,
Therefore ' $T$ ' increases
So, time period of simple pendulum is greater than ' $T$ ' as value of ' $g$ ' is less in mine.

UP CPMT-2005

Gravitation

138378 The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r < R$ depend on $r$ according to relation: $(R =$ radius of earth)

1

g \propto \frac{1}{r^{2}}

2

g \propto \frac{1}{r}

3

g \propto r

4

g \propto r^{2}

[UP CMPT-2002]

Explanation:

C Gravitational force $(F) = mg$

$g = \frac{F}{m}$
$F = \frac{GMm}{R^{3}} r$
$g = \frac{GMmr}{R^{3} m} = \frac{Gm}{R^{3}} r$
$g \propto r$

Gravitation

138379 At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 km$ above the surface of earth? (Radius of earth $= 6400 km$ )

1

1600 km

2

2400 km

3

3200 km

4

4352 km

Explanation:

D Given, $h = 1600 km, R_{e} = 6400 km$ At a height $h$ above the earth's surface
$g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
And below the earth's surface at a depth $x$ ,
$g^{''} = g (1 - \frac{x}{R_{e}})$
According to question
$g^{''} = \frac{g^{'}}{2}$
Then,
$g^{''} = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
From equation (i) and (ii)
$g (1 - \frac{x}{R_{e}}) = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{R_{e}} = \frac{1}{2 {(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{6400} = \frac{1}{2 {(1 + \frac{1600}{6400})}^{2}}$
$1 - \frac{x}{6400} = \frac{8}{25}$
$\frac{x}{6400} = 1 - \frac{8}{25}$
$x = \frac{17}{25} \times 6400$
$x = 4352 km$

UP CPMT-2001

Gravitation

138381 The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

1

9.8 m / s

2

19.6 m / s

3

980 m / s

4

4.9 m / s

Explanation:

D Given $M_{p} = 2 M_{e}, D_{p} = 2 D_{e}$
$g_{p} = \frac{G_{p}}{R_{p}^{2}} = \frac{4 M_{p}}{D_{p}^{2}}$
$= \frac{4 G (2 M_{e})}{{(2 D_{e})}^{2}}$
$= \frac{1}{2} (\frac{4 {GM}_{e}}{D_{e}^{2}})$
$= \frac{1}{2} g_{e}$
$= \frac{1}{2} \times 9.8 m / s$
$g_{p} = 4.9 m / s$

SRMJEE - 2009

Gravitation

138376 What will be the acceleration due to gravity at a depth $d$ , where $g$ is acceleration due to gravity on the surface of earth ?

1

\frac{g}{{[1 + \frac{d}{R}]}^{2}}

2

g [1 - \frac{2 d}{R}]

3

\frac{g}{{[1 - \frac{d}{R}]}^{2}}

4

g [1 - \frac{d}{R}]

Explanation:

D Variation of $g$ with depth

In this figure
$m = mass$
$d = distance below the earth's surface$
We know that on the surface of the earth
$g = \frac{4}{3} π ρ RG [∴ g = \frac{GM}{R^{2}}]$
At a distance (d) below the earth's surface
$g_{d} = \frac{4}{3} π ρ (R - d) G$
$\frac{g}{g_{d}} = \frac{R}{(R - d)}$
$g_{d} = (1 - \frac{d}{R}) g$
When, $d = 0$ then value of $g$ on the surface
And
$g_{d} = g$
When, $d = R$ then value of $g$ at the centre of the earth.
$g_{d} = 0$

UP CPMT-2010

Gravitation

138377 The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be

1 cannot be compared

2 equal to

T

3 less than

T

4 more than

T

Explanation:

D Time Period $(T) = 2 π \sqrt{\frac{l}{g}}$
$T \propto \frac{1}{\sqrt{g}}$
Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $R$ ,
Then,
$g^{'} = g (1 - \frac{h}{R})$
Hence, $g$ decreases,
Therefore ' $T$ ' increases
So, time period of simple pendulum is greater than ' $T$ ' as value of ' $g$ ' is less in mine.

UP CPMT-2005

Gravitation

138378 The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r < R$ depend on $r$ according to relation: $(R =$ radius of earth)

1

g \propto \frac{1}{r^{2}}

2

g \propto \frac{1}{r}

3

g \propto r

4

g \propto r^{2}

[UP CMPT-2002]

Explanation:

C Gravitational force $(F) = mg$

$g = \frac{F}{m}$
$F = \frac{GMm}{R^{3}} r$
$g = \frac{GMmr}{R^{3} m} = \frac{Gm}{R^{3}} r$
$g \propto r$

Gravitation

138379 At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 km$ above the surface of earth? (Radius of earth $= 6400 km$ )

1

1600 km

2

2400 km

3

3200 km

4

4352 km

Explanation:

D Given, $h = 1600 km, R_{e} = 6400 km$ At a height $h$ above the earth's surface
$g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
And below the earth's surface at a depth $x$ ,
$g^{''} = g (1 - \frac{x}{R_{e}})$
According to question
$g^{''} = \frac{g^{'}}{2}$
Then,
$g^{''} = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
From equation (i) and (ii)
$g (1 - \frac{x}{R_{e}}) = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{R_{e}} = \frac{1}{2 {(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{6400} = \frac{1}{2 {(1 + \frac{1600}{6400})}^{2}}$
$1 - \frac{x}{6400} = \frac{8}{25}$
$\frac{x}{6400} = 1 - \frac{8}{25}$
$x = \frac{17}{25} \times 6400$
$x = 4352 km$

UP CPMT-2001

Gravitation

138381 The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

1

9.8 m / s

2

19.6 m / s

3

980 m / s

4

4.9 m / s

Explanation:

D Given $M_{p} = 2 M_{e}, D_{p} = 2 D_{e}$
$g_{p} = \frac{G_{p}}{R_{p}^{2}} = \frac{4 M_{p}}{D_{p}^{2}}$
$= \frac{4 G (2 M_{e})}{{(2 D_{e})}^{2}}$
$= \frac{1}{2} (\frac{4 {GM}_{e}}{D_{e}^{2}})$
$= \frac{1}{2} g_{e}$
$= \frac{1}{2} \times 9.8 m / s$
$g_{p} = 4.9 m / s$

SRMJEE - 2009

Gravitation

138376 What will be the acceleration due to gravity at a depth $d$ , where $g$ is acceleration due to gravity on the surface of earth ?

1

\frac{g}{{[1 + \frac{d}{R}]}^{2}}

2

g [1 - \frac{2 d}{R}]

3

\frac{g}{{[1 - \frac{d}{R}]}^{2}}

4

g [1 - \frac{d}{R}]

Explanation:

D Variation of $g$ with depth

In this figure
$m = mass$
$d = distance below the earth's surface$
We know that on the surface of the earth
$g = \frac{4}{3} π ρ RG [∴ g = \frac{GM}{R^{2}}]$
At a distance (d) below the earth's surface
$g_{d} = \frac{4}{3} π ρ (R - d) G$
$\frac{g}{g_{d}} = \frac{R}{(R - d)}$
$g_{d} = (1 - \frac{d}{R}) g$
When, $d = 0$ then value of $g$ on the surface
And
$g_{d} = g$
When, $d = R$ then value of $g$ at the centre of the earth.
$g_{d} = 0$

UP CPMT-2010

Gravitation

138377 The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be

1 cannot be compared

2 equal to

T

3 less than

T

4 more than

T

Explanation:

D Time Period $(T) = 2 π \sqrt{\frac{l}{g}}$
$T \propto \frac{1}{\sqrt{g}}$
Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $R$ ,
Then,
$g^{'} = g (1 - \frac{h}{R})$
Hence, $g$ decreases,
Therefore ' $T$ ' increases
So, time period of simple pendulum is greater than ' $T$ ' as value of ' $g$ ' is less in mine.

UP CPMT-2005

Gravitation

138378 The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r < R$ depend on $r$ according to relation: $(R =$ radius of earth)

1

g \propto \frac{1}{r^{2}}

2

g \propto \frac{1}{r}

3

g \propto r

4

g \propto r^{2}

[UP CMPT-2002]

Explanation:

C Gravitational force $(F) = mg$

$g = \frac{F}{m}$
$F = \frac{GMm}{R^{3}} r$
$g = \frac{GMmr}{R^{3} m} = \frac{Gm}{R^{3}} r$
$g \propto r$

Gravitation

138379 At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 km$ above the surface of earth? (Radius of earth $= 6400 km$ )

1

1600 km

2

2400 km

3

3200 km

4

4352 km

Explanation:

D Given, $h = 1600 km, R_{e} = 6400 km$ At a height $h$ above the earth's surface
$g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
And below the earth's surface at a depth $x$ ,
$g^{''} = g (1 - \frac{x}{R_{e}})$
According to question
$g^{''} = \frac{g^{'}}{2}$
Then,
$g^{''} = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
From equation (i) and (ii)
$g (1 - \frac{x}{R_{e}}) = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{R_{e}} = \frac{1}{2 {(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{6400} = \frac{1}{2 {(1 + \frac{1600}{6400})}^{2}}$
$1 - \frac{x}{6400} = \frac{8}{25}$
$\frac{x}{6400} = 1 - \frac{8}{25}$
$x = \frac{17}{25} \times 6400$
$x = 4352 km$

UP CPMT-2001

Gravitation

138381 The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

1

9.8 m / s

2

19.6 m / s

3

980 m / s

4

4.9 m / s

Explanation:

D Given $M_{p} = 2 M_{e}, D_{p} = 2 D_{e}$
$g_{p} = \frac{G_{p}}{R_{p}^{2}} = \frac{4 M_{p}}{D_{p}^{2}}$
$= \frac{4 G (2 M_{e})}{{(2 D_{e})}^{2}}$
$= \frac{1}{2} (\frac{4 {GM}_{e}}{D_{e}^{2}})$
$= \frac{1}{2} g_{e}$
$= \frac{1}{2} \times 9.8 m / s$
$g_{p} = 4.9 m / s$

SRMJEE - 2009

Gravitation

138376 What will be the acceleration due to gravity at a depth $d$ , where $g$ is acceleration due to gravity on the surface of earth ?

1

\frac{g}{{[1 + \frac{d}{R}]}^{2}}

2

g [1 - \frac{2 d}{R}]

3

\frac{g}{{[1 - \frac{d}{R}]}^{2}}

4

g [1 - \frac{d}{R}]

Explanation:

D Variation of $g$ with depth

In this figure
$m = mass$
$d = distance below the earth's surface$
We know that on the surface of the earth
$g = \frac{4}{3} π ρ RG [∴ g = \frac{GM}{R^{2}}]$
At a distance (d) below the earth's surface
$g_{d} = \frac{4}{3} π ρ (R - d) G$
$\frac{g}{g_{d}} = \frac{R}{(R - d)}$
$g_{d} = (1 - \frac{d}{R}) g$
When, $d = 0$ then value of $g$ on the surface
And
$g_{d} = g$
When, $d = R$ then value of $g$ at the centre of the earth.
$g_{d} = 0$

UP CPMT-2010

Gravitation

138377 The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be

1 cannot be compared

2 equal to

T

3 less than

T

4 more than

T

Explanation:

D Time Period $(T) = 2 π \sqrt{\frac{l}{g}}$
$T \propto \frac{1}{\sqrt{g}}$
Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $R$ ,
Then,
$g^{'} = g (1 - \frac{h}{R})$
Hence, $g$ decreases,
Therefore ' $T$ ' increases
So, time period of simple pendulum is greater than ' $T$ ' as value of ' $g$ ' is less in mine.

UP CPMT-2005

Gravitation

138378 The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r < R$ depend on $r$ according to relation: $(R =$ radius of earth)

1

g \propto \frac{1}{r^{2}}

2

g \propto \frac{1}{r}

3

g \propto r

4

g \propto r^{2}

[UP CMPT-2002]

Explanation:

C Gravitational force $(F) = mg$

$g = \frac{F}{m}$
$F = \frac{GMm}{R^{3}} r$
$g = \frac{GMmr}{R^{3} m} = \frac{Gm}{R^{3}} r$
$g \propto r$

Gravitation

138379 At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 km$ above the surface of earth? (Radius of earth $= 6400 km$ )

1

1600 km

2

2400 km

3

3200 km

4

4352 km

Explanation:

D Given, $h = 1600 km, R_{e} = 6400 km$ At a height $h$ above the earth's surface
$g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
And below the earth's surface at a depth $x$ ,
$g^{''} = g (1 - \frac{x}{R_{e}})$
According to question
$g^{''} = \frac{g^{'}}{2}$
Then,
$g^{''} = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
From equation (i) and (ii)
$g (1 - \frac{x}{R_{e}}) = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{R_{e}} = \frac{1}{2 {(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{6400} = \frac{1}{2 {(1 + \frac{1600}{6400})}^{2}}$
$1 - \frac{x}{6400} = \frac{8}{25}$
$\frac{x}{6400} = 1 - \frac{8}{25}$
$x = \frac{17}{25} \times 6400$
$x = 4352 km$

UP CPMT-2001

Gravitation

138381 The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

1

9.8 m / s

2

19.6 m / s

3

980 m / s

4

4.9 m / s

Explanation:

D Given $M_{p} = 2 M_{e}, D_{p} = 2 D_{e}$
$g_{p} = \frac{G_{p}}{R_{p}^{2}} = \frac{4 M_{p}}{D_{p}^{2}}$
$= \frac{4 G (2 M_{e})}{{(2 D_{e})}^{2}}$
$= \frac{1}{2} (\frac{4 {GM}_{e}}{D_{e}^{2}})$
$= \frac{1}{2} g_{e}$
$= \frac{1}{2} \times 9.8 m / s$
$g_{p} = 4.9 m / s$

SRMJEE - 2009

Gravitation

138376 What will be the acceleration due to gravity at a depth $d$ , where $g$ is acceleration due to gravity on the surface of earth ?

1

\frac{g}{{[1 + \frac{d}{R}]}^{2}}

2

g [1 - \frac{2 d}{R}]

3

\frac{g}{{[1 - \frac{d}{R}]}^{2}}

4

g [1 - \frac{d}{R}]

Explanation:

D Variation of $g$ with depth

In this figure
$m = mass$
$d = distance below the earth's surface$
We know that on the surface of the earth
$g = \frac{4}{3} π ρ RG [∴ g = \frac{GM}{R^{2}}]$
At a distance (d) below the earth's surface
$g_{d} = \frac{4}{3} π ρ (R - d) G$
$\frac{g}{g_{d}} = \frac{R}{(R - d)}$
$g_{d} = (1 - \frac{d}{R}) g$
When, $d = 0$ then value of $g$ on the surface
And
$g_{d} = g$
When, $d = R$ then value of $g$ at the centre of the earth.
$g_{d} = 0$

UP CPMT-2010

Gravitation

138377 The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be

1 cannot be compared

2 equal to

T

3 less than

T

4 more than

T

Explanation:

D Time Period $(T) = 2 π \sqrt{\frac{l}{g}}$
$T \propto \frac{1}{\sqrt{g}}$
Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $R$ ,
Then,
$g^{'} = g (1 - \frac{h}{R})$
Hence, $g$ decreases,
Therefore ' $T$ ' increases
So, time period of simple pendulum is greater than ' $T$ ' as value of ' $g$ ' is less in mine.

UP CPMT-2005

Gravitation

138378 The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r < R$ depend on $r$ according to relation: $(R =$ radius of earth)

1

g \propto \frac{1}{r^{2}}

2

g \propto \frac{1}{r}

3

g \propto r

4

g \propto r^{2}

[UP CMPT-2002]

Explanation:

C Gravitational force $(F) = mg$

$g = \frac{F}{m}$
$F = \frac{GMm}{R^{3}} r$
$g = \frac{GMmr}{R^{3} m} = \frac{Gm}{R^{3}} r$
$g \propto r$

Gravitation

138379 At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 km$ above the surface of earth? (Radius of earth $= 6400 km$ )

1

1600 km

2

2400 km

3

3200 km

4

4352 km

Explanation:

D Given, $h = 1600 km, R_{e} = 6400 km$ At a height $h$ above the earth's surface
$g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
And below the earth's surface at a depth $x$ ,
$g^{''} = g (1 - \frac{x}{R_{e}})$
According to question
$g^{''} = \frac{g^{'}}{2}$
Then,
$g^{''} = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
From equation (i) and (ii)
$g (1 - \frac{x}{R_{e}}) = \frac{1}{2} \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{R_{e}} = \frac{1}{2 {(1 + \frac{h}{R_{e}})}^{2}}$
$1 - \frac{x}{6400} = \frac{1}{2 {(1 + \frac{1600}{6400})}^{2}}$
$1 - \frac{x}{6400} = \frac{8}{25}$
$\frac{x}{6400} = 1 - \frac{8}{25}$
$x = \frac{17}{25} \times 6400$
$x = 4352 km$

UP CPMT-2001

Gravitation

138381 The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

1

9.8 m / s

2

19.6 m / s

3

980 m / s

4

4.9 m / s

Explanation:

D Given $M_{p} = 2 M_{e}, D_{p} = 2 D_{e}$
$g_{p} = \frac{G_{p}}{R_{p}^{2}} = \frac{4 M_{p}}{D_{p}^{2}}$
$= \frac{4 G (2 M_{e})}{{(2 D_{e})}^{2}}$
$= \frac{1}{2} (\frac{4 {GM}_{e}}{D_{e}^{2}})$
$= \frac{1}{2} g_{e}$
$= \frac{1}{2} \times 9.8 m / s$
$g_{p} = 4.9 m / s$

SRMJEE - 2009

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