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PHXI08:GRAVITATION

359683 A body weight $W$ , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be

1

\frac{W}{100}

2

\frac{W}{91}

3

\frac{W}{9}

4

\frac{W}{3}

Explanation:

As, $W = \frac{G M_{e}}{R_{e}^{2}} \times m$
Weight of the body at height $h$ given by
$W^{'} = \frac{G M_{e} m}{{(R_{e} + h)}^{2}}$
$W^{'} = \frac{G M_{e}}{{(9 R_{e} + R_{e})}^{2}} \times m = \frac{G M_{e} m}{{(10 R_{e})}^{2}} = \frac{G M_{e} m}{100 R_{e}^{2}}$
supporting img
$W^{'} = \frac{1}{100} W$
So, correct option is (1).

JEE - 2023

PHXI08:GRAVITATION

359684 The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 %$ of its value at the surface is $(R$ is the radius of the earth)

1

8 R

2

9 R

3

10 R

4

20 R

Explanation:

Acceleration due to gravity at a height $h$ above the earth's surface is $g = \frac{G M}{{(R + h)}^{2}}$ where, $G =$ Universal gravitational constant
$R =$ Radius of the earth
$g^{'} = \frac{G M}{R^{2} {(1 + \frac{h}{R})}^{2}} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
According to given problem, $g^{'} = \frac{g}{100}$
$\frac{g}{100} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 9 R$

PHXI08:GRAVITATION

359685 Assertion :
Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height.
Reason :
The path of a projectile is independent of the gravitational force of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

PHXI08:GRAVITATION

359686 The fractional change in the value of free-fall acceleration $g$ for a particle when it is lifted from the surface to an elevation $h (h << R)$ is

1

h / R

2

- (2 h / R)

3

2 h / R

4 none of these

Explanation:

$g = \frac{G M}{R^{2}} (1)$
$\frac{d g}{d R} = \frac{- 2 G M}{R^{3}}$ putting $d R = h$ , we obtain
$\Rightarrow \frac{d g}{h} = \frac{- 2 G M}{R^{2}} \cdot \frac{1}{R} (2)$
From (1) and (2) $\Rightarrow \frac{d g}{g} = - 2 (\frac{h}{R})$
$\Rightarrow$ Change is $- v e$ . That means $g$ decreases.

PHXI08:GRAVITATION

359687 Two equal masses $m$ and $m$ are hung from balance whose scale pans differ in vertical height by $h$ . calculate the error in weighing, if any, in terms of density of earth $ρ$ .
supporting img

1

\frac{8}{3} π ρ G m h

2

\frac{4}{3} π ρ G m^{2} h

3

\frac{2}{3} π ρ R^{3} G m

4

\frac{8}{3} π ρ R^{3} G m

Explanation:

During normal weighing method we maintain equal lengths for both pans. In this case the acceleration due to gravity is same for both bodies. In this case pan heights are different and hence $g$ is different at both locations. As with height $g$ varies as
$g^{'} = \frac{g}{[1 + h / R]^{2}} = g [1 - \frac{2 h}{R}]$
supporting img
and in according with figure $h_{1} > h_{2}, W_{1}$ will be lesser than $W_{2}$ and
i.e., $W_{2} - W_{1} = m g_{2} - m g_{1} = 2 m g [\frac{h_{1}}{R} - \frac{h_{2}}{R}]$ or $W_{2} - W_{1} = 2 m \frac{G M}{R^{2}} \frac{h}{R}$
$[As g = \frac{G M}{R^{2}} and (h_{1} - h_{2}) = h]$
or $W_{2} - W_{1} = \frac{2 m h G}{R^{3}} (\frac{4}{3} π R^{2} ρ)$
$\begin{aligned} = \frac{8}{3} π ρ G m h \\ [as M = \frac{4}{3} π R^{3} ρ] \end{aligned}$

PHXI08:GRAVITATION

359683 A body weight $W$ , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be

1

\frac{W}{100}

2

\frac{W}{91}

3

\frac{W}{9}

4

\frac{W}{3}

Explanation:

As, $W = \frac{G M_{e}}{R_{e}^{2}} \times m$
Weight of the body at height $h$ given by
$W^{'} = \frac{G M_{e} m}{{(R_{e} + h)}^{2}}$
$W^{'} = \frac{G M_{e}}{{(9 R_{e} + R_{e})}^{2}} \times m = \frac{G M_{e} m}{{(10 R_{e})}^{2}} = \frac{G M_{e} m}{100 R_{e}^{2}}$
supporting img
$W^{'} = \frac{1}{100} W$
So, correct option is (1).

JEE - 2023

PHXI08:GRAVITATION

359684 The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 %$ of its value at the surface is $(R$ is the radius of the earth)

1

8 R

2

9 R

3

10 R

4

20 R

Explanation:

Acceleration due to gravity at a height $h$ above the earth's surface is $g = \frac{G M}{{(R + h)}^{2}}$ where, $G =$ Universal gravitational constant
$R =$ Radius of the earth
$g^{'} = \frac{G M}{R^{2} {(1 + \frac{h}{R})}^{2}} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
According to given problem, $g^{'} = \frac{g}{100}$
$\frac{g}{100} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 9 R$

PHXI08:GRAVITATION

359685 Assertion :
Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height.
Reason :
The path of a projectile is independent of the gravitational force of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

PHXI08:GRAVITATION

359686 The fractional change in the value of free-fall acceleration $g$ for a particle when it is lifted from the surface to an elevation $h (h << R)$ is

1

h / R

2

- (2 h / R)

3

2 h / R

4 none of these

Explanation:

$g = \frac{G M}{R^{2}} (1)$
$\frac{d g}{d R} = \frac{- 2 G M}{R^{3}}$ putting $d R = h$ , we obtain
$\Rightarrow \frac{d g}{h} = \frac{- 2 G M}{R^{2}} \cdot \frac{1}{R} (2)$
From (1) and (2) $\Rightarrow \frac{d g}{g} = - 2 (\frac{h}{R})$
$\Rightarrow$ Change is $- v e$ . That means $g$ decreases.

PHXI08:GRAVITATION

359687 Two equal masses $m$ and $m$ are hung from balance whose scale pans differ in vertical height by $h$ . calculate the error in weighing, if any, in terms of density of earth $ρ$ .
supporting img

1

\frac{8}{3} π ρ G m h

2

\frac{4}{3} π ρ G m^{2} h

3

\frac{2}{3} π ρ R^{3} G m

4

\frac{8}{3} π ρ R^{3} G m

Explanation:

During normal weighing method we maintain equal lengths for both pans. In this case the acceleration due to gravity is same for both bodies. In this case pan heights are different and hence $g$ is different at both locations. As with height $g$ varies as
$g^{'} = \frac{g}{[1 + h / R]^{2}} = g [1 - \frac{2 h}{R}]$
supporting img
and in according with figure $h_{1} > h_{2}, W_{1}$ will be lesser than $W_{2}$ and
i.e., $W_{2} - W_{1} = m g_{2} - m g_{1} = 2 m g [\frac{h_{1}}{R} - \frac{h_{2}}{R}]$ or $W_{2} - W_{1} = 2 m \frac{G M}{R^{2}} \frac{h}{R}$
$[As g = \frac{G M}{R^{2}} and (h_{1} - h_{2}) = h]$
or $W_{2} - W_{1} = \frac{2 m h G}{R^{3}} (\frac{4}{3} π R^{2} ρ)$
$\begin{aligned} = \frac{8}{3} π ρ G m h \\ [as M = \frac{4}{3} π R^{3} ρ] \end{aligned}$

PHXI08:GRAVITATION

359683 A body weight $W$ , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be

1

\frac{W}{100}

2

\frac{W}{91}

3

\frac{W}{9}

4

\frac{W}{3}

Explanation:

As, $W = \frac{G M_{e}}{R_{e}^{2}} \times m$
Weight of the body at height $h$ given by
$W^{'} = \frac{G M_{e} m}{{(R_{e} + h)}^{2}}$
$W^{'} = \frac{G M_{e}}{{(9 R_{e} + R_{e})}^{2}} \times m = \frac{G M_{e} m}{{(10 R_{e})}^{2}} = \frac{G M_{e} m}{100 R_{e}^{2}}$
supporting img
$W^{'} = \frac{1}{100} W$
So, correct option is (1).

JEE - 2023

PHXI08:GRAVITATION

359684 The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 %$ of its value at the surface is $(R$ is the radius of the earth)

1

8 R

2

9 R

3

10 R

4

20 R

Explanation:

Acceleration due to gravity at a height $h$ above the earth's surface is $g = \frac{G M}{{(R + h)}^{2}}$ where, $G =$ Universal gravitational constant
$R =$ Radius of the earth
$g^{'} = \frac{G M}{R^{2} {(1 + \frac{h}{R})}^{2}} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
According to given problem, $g^{'} = \frac{g}{100}$
$\frac{g}{100} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 9 R$

PHXI08:GRAVITATION

359685 Assertion :
Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height.
Reason :
The path of a projectile is independent of the gravitational force of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

PHXI08:GRAVITATION

359686 The fractional change in the value of free-fall acceleration $g$ for a particle when it is lifted from the surface to an elevation $h (h << R)$ is

1

h / R

2

- (2 h / R)

3

2 h / R

4 none of these

Explanation:

$g = \frac{G M}{R^{2}} (1)$
$\frac{d g}{d R} = \frac{- 2 G M}{R^{3}}$ putting $d R = h$ , we obtain
$\Rightarrow \frac{d g}{h} = \frac{- 2 G M}{R^{2}} \cdot \frac{1}{R} (2)$
From (1) and (2) $\Rightarrow \frac{d g}{g} = - 2 (\frac{h}{R})$
$\Rightarrow$ Change is $- v e$ . That means $g$ decreases.

PHXI08:GRAVITATION

359687 Two equal masses $m$ and $m$ are hung from balance whose scale pans differ in vertical height by $h$ . calculate the error in weighing, if any, in terms of density of earth $ρ$ .
supporting img

1

\frac{8}{3} π ρ G m h

2

\frac{4}{3} π ρ G m^{2} h

3

\frac{2}{3} π ρ R^{3} G m

4

\frac{8}{3} π ρ R^{3} G m

Explanation:

During normal weighing method we maintain equal lengths for both pans. In this case the acceleration due to gravity is same for both bodies. In this case pan heights are different and hence $g$ is different at both locations. As with height $g$ varies as
$g^{'} = \frac{g}{[1 + h / R]^{2}} = g [1 - \frac{2 h}{R}]$
supporting img
and in according with figure $h_{1} > h_{2}, W_{1}$ will be lesser than $W_{2}$ and
i.e., $W_{2} - W_{1} = m g_{2} - m g_{1} = 2 m g [\frac{h_{1}}{R} - \frac{h_{2}}{R}]$ or $W_{2} - W_{1} = 2 m \frac{G M}{R^{2}} \frac{h}{R}$
$[As g = \frac{G M}{R^{2}} and (h_{1} - h_{2}) = h]$
or $W_{2} - W_{1} = \frac{2 m h G}{R^{3}} (\frac{4}{3} π R^{2} ρ)$
$\begin{aligned} = \frac{8}{3} π ρ G m h \\ [as M = \frac{4}{3} π R^{3} ρ] \end{aligned}$

PHXI08:GRAVITATION

359683 A body weight $W$ , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be

1

\frac{W}{100}

2

\frac{W}{91}

3

\frac{W}{9}

4

\frac{W}{3}

Explanation:

As, $W = \frac{G M_{e}}{R_{e}^{2}} \times m$
Weight of the body at height $h$ given by
$W^{'} = \frac{G M_{e} m}{{(R_{e} + h)}^{2}}$
$W^{'} = \frac{G M_{e}}{{(9 R_{e} + R_{e})}^{2}} \times m = \frac{G M_{e} m}{{(10 R_{e})}^{2}} = \frac{G M_{e} m}{100 R_{e}^{2}}$
supporting img
$W^{'} = \frac{1}{100} W$
So, correct option is (1).

JEE - 2023

PHXI08:GRAVITATION

359684 The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 %$ of its value at the surface is $(R$ is the radius of the earth)

1

8 R

2

9 R

3

10 R

4

20 R

Explanation:

Acceleration due to gravity at a height $h$ above the earth's surface is $g = \frac{G M}{{(R + h)}^{2}}$ where, $G =$ Universal gravitational constant
$R =$ Radius of the earth
$g^{'} = \frac{G M}{R^{2} {(1 + \frac{h}{R})}^{2}} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
According to given problem, $g^{'} = \frac{g}{100}$
$\frac{g}{100} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 9 R$

PHXI08:GRAVITATION

359685 Assertion :
Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height.
Reason :
The path of a projectile is independent of the gravitational force of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

PHXI08:GRAVITATION

359686 The fractional change in the value of free-fall acceleration $g$ for a particle when it is lifted from the surface to an elevation $h (h << R)$ is

1

h / R

2

- (2 h / R)

3

2 h / R

4 none of these

Explanation:

$g = \frac{G M}{R^{2}} (1)$
$\frac{d g}{d R} = \frac{- 2 G M}{R^{3}}$ putting $d R = h$ , we obtain
$\Rightarrow \frac{d g}{h} = \frac{- 2 G M}{R^{2}} \cdot \frac{1}{R} (2)$
From (1) and (2) $\Rightarrow \frac{d g}{g} = - 2 (\frac{h}{R})$
$\Rightarrow$ Change is $- v e$ . That means $g$ decreases.

PHXI08:GRAVITATION

359687 Two equal masses $m$ and $m$ are hung from balance whose scale pans differ in vertical height by $h$ . calculate the error in weighing, if any, in terms of density of earth $ρ$ .
supporting img

1

\frac{8}{3} π ρ G m h

2

\frac{4}{3} π ρ G m^{2} h

3

\frac{2}{3} π ρ R^{3} G m

4

\frac{8}{3} π ρ R^{3} G m

Explanation:

During normal weighing method we maintain equal lengths for both pans. In this case the acceleration due to gravity is same for both bodies. In this case pan heights are different and hence $g$ is different at both locations. As with height $g$ varies as
$g^{'} = \frac{g}{[1 + h / R]^{2}} = g [1 - \frac{2 h}{R}]$
supporting img
and in according with figure $h_{1} > h_{2}, W_{1}$ will be lesser than $W_{2}$ and
i.e., $W_{2} - W_{1} = m g_{2} - m g_{1} = 2 m g [\frac{h_{1}}{R} - \frac{h_{2}}{R}]$ or $W_{2} - W_{1} = 2 m \frac{G M}{R^{2}} \frac{h}{R}$
$[As g = \frac{G M}{R^{2}} and (h_{1} - h_{2}) = h]$
or $W_{2} - W_{1} = \frac{2 m h G}{R^{3}} (\frac{4}{3} π R^{2} ρ)$
$\begin{aligned} = \frac{8}{3} π ρ G m h \\ [as M = \frac{4}{3} π R^{3} ρ] \end{aligned}$

PHXI08:GRAVITATION

359683 A body weight $W$ , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be

1

\frac{W}{100}

2

\frac{W}{91}

3

\frac{W}{9}

4

\frac{W}{3}

Explanation:

As, $W = \frac{G M_{e}}{R_{e}^{2}} \times m$
Weight of the body at height $h$ given by
$W^{'} = \frac{G M_{e} m}{{(R_{e} + h)}^{2}}$
$W^{'} = \frac{G M_{e}}{{(9 R_{e} + R_{e})}^{2}} \times m = \frac{G M_{e} m}{{(10 R_{e})}^{2}} = \frac{G M_{e} m}{100 R_{e}^{2}}$
supporting img
$W^{'} = \frac{1}{100} W$
So, correct option is (1).

JEE - 2023

PHXI08:GRAVITATION

359684 The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 %$ of its value at the surface is $(R$ is the radius of the earth)

1

8 R

2

9 R

3

10 R

4

20 R

Explanation:

Acceleration due to gravity at a height $h$ above the earth's surface is $g = \frac{G M}{{(R + h)}^{2}}$ where, $G =$ Universal gravitational constant
$R =$ Radius of the earth
$g^{'} = \frac{G M}{R^{2} {(1 + \frac{h}{R})}^{2}} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
According to given problem, $g^{'} = \frac{g}{100}$
$\frac{g}{100} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 9 R$

PHXI08:GRAVITATION

359685 Assertion :
Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height.
Reason :
The path of a projectile is independent of the gravitational force of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

PHXI08:GRAVITATION

359686 The fractional change in the value of free-fall acceleration $g$ for a particle when it is lifted from the surface to an elevation $h (h << R)$ is

1

h / R

2

- (2 h / R)

3

2 h / R

4 none of these

Explanation:

$g = \frac{G M}{R^{2}} (1)$
$\frac{d g}{d R} = \frac{- 2 G M}{R^{3}}$ putting $d R = h$ , we obtain
$\Rightarrow \frac{d g}{h} = \frac{- 2 G M}{R^{2}} \cdot \frac{1}{R} (2)$
From (1) and (2) $\Rightarrow \frac{d g}{g} = - 2 (\frac{h}{R})$
$\Rightarrow$ Change is $- v e$ . That means $g$ decreases.

PHXI08:GRAVITATION

359687 Two equal masses $m$ and $m$ are hung from balance whose scale pans differ in vertical height by $h$ . calculate the error in weighing, if any, in terms of density of earth $ρ$ .
supporting img

1

\frac{8}{3} π ρ G m h

2

\frac{4}{3} π ρ G m^{2} h

3

\frac{2}{3} π ρ R^{3} G m

4

\frac{8}{3} π ρ R^{3} G m

Explanation:

During normal weighing method we maintain equal lengths for both pans. In this case the acceleration due to gravity is same for both bodies. In this case pan heights are different and hence $g$ is different at both locations. As with height $g$ varies as
$g^{'} = \frac{g}{[1 + h / R]^{2}} = g [1 - \frac{2 h}{R}]$
supporting img
and in according with figure $h_{1} > h_{2}, W_{1}$ will be lesser than $W_{2}$ and
i.e., $W_{2} - W_{1} = m g_{2} - m g_{1} = 2 m g [\frac{h_{1}}{R} - \frac{h_{2}}{R}]$ or $W_{2} - W_{1} = 2 m \frac{G M}{R^{2}} \frac{h}{R}$
$[As g = \frac{G M}{R^{2}} and (h_{1} - h_{2}) = h]$
or $W_{2} - W_{1} = \frac{2 m h G}{R^{3}} (\frac{4}{3} π R^{2} ρ)$
$\begin{aligned} = \frac{8}{3} π ρ G m h \\ [as M = \frac{4}{3} π R^{3} ρ] \end{aligned}$

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