QBManager

Gravitation

138270 Suppose the force of gravitation between two bodies of equal masses is $F$ . If each mass is doubled keeping the distance of separation between them unchanged, the force would become

1

F

2

2 F

3

4 F

4

\frac{1}{4} F

Explanation:

C Two bodies are of equal mass.
So, $m_{1} = m_{2} = m$
$m_{1} =$ mass of $1^{st}$ body, $m_{2} =$ mass of $2^{nd}$ body and $r =$ distance between the two bodies
Then, $F = G \frac{m_{1} m_{2}}{r^{2}}$
$F = \frac{{Gm}^{2}}{r^{2}}$
Now masses are doubled and distance is the same
$m_{1} = m_{2} = 2 m$
The magnitude of the gravitational force $F$ using equation (i)
$F = G \frac{2 m \times 2 m}{r^{2}}$
$F^{1} = 4 \times \frac{{Gm}^{2}}{r^{2}} = 4 F$

NDA (II) 2019

Gravitation

138271 Which one of the following statements is true for the relation $F = \frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)

1 The quantity

G

depends on the local value of

g

, acceleration due to gravity

2 The quantity

G

is greatest at the surface of the Earth

3 The quantity G is used only when earth is one of the two masses

4 The quantity

G

is a universal constant

Explanation:

D Gravitational Force $(F) = G \frac{m_{1} m_{2}}{r^{2}}$
Where, $m_{1} =$ mass of first body
$m_{2} =$ mass of second body
$r =$ distance between two body
$G =$ Gravitational constant
$G$ is the universal gravitational constant which remain constant at all places in the universe $G$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart.
The value of $G = 6.67 \times 10^{- 11} {Nm}^{2} / {kg}^{2}$

NDA (I) 2017

Gravitation

138273 Three particles, two with masses $m$ and one with mass $M$ , might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$ , least to greatest.

1

1, 2, 3, 4

2

2, 1, 3, 4

3

2, 1, 4, 3

4 2, 3, 4, 1

Explanation:

B From figure (i),

$∵ F_{net} = F_{1} + F_{2}$
$= \frac{GMm}{d^{2}} + \frac{GMm}{(2 d)^{2}}$
$= \frac{GMm}{d^{2}} (1 + \frac{1}{4})$
$= \frac{5}{4} \frac{GMm}{d^{2}} = 1.25 \frac{GMm}{d^{2}}$
From figure (ii),

$∵ F_{net} = F_{4} - F_{3}$
$= \frac{GMm}{d^{2}} - \frac{GMm}{d^{2}} = 0$
From figure (iii),

$∵ F_{net} = \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos θ}$
$= \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos 90^{\circ}} (∵ \cos 90^{\circ} = 0)$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2}}$
$F_{net} = \sqrt{2} \frac{GMm}{d^{2}} = 1.414 \frac{GMm}{d^{2}}$
From equation (iv),

$∵ F_{net} = \sqrt{F_{7}^{2} + F_{8}^{2} + 2 F_{7} F_{8} \cos θ}$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2} + 2 \frac{GMm}{d^{2}} \times \frac{GMm}{d^{2}} \cos θ}$
$= \frac{GMm}{d^{2}} \sqrt{1 + 1 + 2 \cos θ}$
$= \frac{GMm}{d^{2}} (\sqrt{2 (1 + \cos θ)})$
$= \sqrt{2} \frac{GMm}{d^{2}} \sqrt{1 + \cos θ} = \sqrt{2} \frac{GMm}{d^{2}} \sqrt{2} \cos \frac{θ}{2}$
$= 2 \frac{GMm}{d^{2}} \cos \frac{θ}{2}$
For, $0 < θ < 90^{\circ}$ then,
$∴ F_{net} > \sqrt{2} \frac{GMm}{d^{2}}$
Thus, the rank of gravitational force from least to greatest, $2 < 1 < 3 < 4$

AMU-2018

Gravitation

138274 The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order.

1 (i), (iii) = (iv), (ii)

2 (i) = (iii), (ii), (iv)

3 (i), (ii), (iii), (iv)

4 (iv), (iii), (ii), (ii)

Explanation:

A We know that,
Gravitational force $(F) = \frac{{Gm}_{1} m_{2}}{r^{2}}$
From figure (i),
$F_{1} = \frac{Gmm}{d^{2}} + \frac{Gmm}{D^{2}} = {Gm}^{2} (\frac{1}{d^{2}} + \frac{1}{D^{2}})$
From figure (ii),
$F_{2} = \frac{Gmm}{d^{2}} - \frac{Gmm}{(D - d)^{2}} = {Gm}^{2} (\frac{1}{d^{2}} - \frac{1}{D^{2}})$
From figure (iii),
$F_{3} = \sqrt{{(\frac{G m m}{d^{2}})}^{2} + {(\frac{G m m}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} + {(\frac{1}{(D)^{2}})}^{2}}$
From figure (iv),
$F_{4} = \sqrt{{(\frac{Gmm}{d^{2}})}^{2} - {(\frac{Gmm}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} - {(\frac{1}{D^{2}})}^{2}}$
Hence, the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order-
$(i), (iii) = (iv), (ii)$

AMU-2013

Gravitation

138270 Suppose the force of gravitation between two bodies of equal masses is $F$ . If each mass is doubled keeping the distance of separation between them unchanged, the force would become

1

F

2

2 F

3

4 F

4

\frac{1}{4} F

Explanation:

C Two bodies are of equal mass.
So, $m_{1} = m_{2} = m$
$m_{1} =$ mass of $1^{st}$ body, $m_{2} =$ mass of $2^{nd}$ body and $r =$ distance between the two bodies
Then, $F = G \frac{m_{1} m_{2}}{r^{2}}$
$F = \frac{{Gm}^{2}}{r^{2}}$
Now masses are doubled and distance is the same
$m_{1} = m_{2} = 2 m$
The magnitude of the gravitational force $F$ using equation (i)
$F = G \frac{2 m \times 2 m}{r^{2}}$
$F^{1} = 4 \times \frac{{Gm}^{2}}{r^{2}} = 4 F$

NDA (II) 2019

Gravitation

138271 Which one of the following statements is true for the relation $F = \frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)

1 The quantity

G

depends on the local value of

g

, acceleration due to gravity

2 The quantity

G

is greatest at the surface of the Earth

3 The quantity G is used only when earth is one of the two masses

4 The quantity

G

is a universal constant

Explanation:

D Gravitational Force $(F) = G \frac{m_{1} m_{2}}{r^{2}}$
Where, $m_{1} =$ mass of first body
$m_{2} =$ mass of second body
$r =$ distance between two body
$G =$ Gravitational constant
$G$ is the universal gravitational constant which remain constant at all places in the universe $G$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart.
The value of $G = 6.67 \times 10^{- 11} {Nm}^{2} / {kg}^{2}$

NDA (I) 2017

Gravitation

138273 Three particles, two with masses $m$ and one with mass $M$ , might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$ , least to greatest.

1

1, 2, 3, 4

2

2, 1, 3, 4

3

2, 1, 4, 3

4 2, 3, 4, 1

Explanation:

B From figure (i),

$∵ F_{net} = F_{1} + F_{2}$
$= \frac{GMm}{d^{2}} + \frac{GMm}{(2 d)^{2}}$
$= \frac{GMm}{d^{2}} (1 + \frac{1}{4})$
$= \frac{5}{4} \frac{GMm}{d^{2}} = 1.25 \frac{GMm}{d^{2}}$
From figure (ii),

$∵ F_{net} = F_{4} - F_{3}$
$= \frac{GMm}{d^{2}} - \frac{GMm}{d^{2}} = 0$
From figure (iii),

$∵ F_{net} = \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos θ}$
$= \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos 90^{\circ}} (∵ \cos 90^{\circ} = 0)$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2}}$
$F_{net} = \sqrt{2} \frac{GMm}{d^{2}} = 1.414 \frac{GMm}{d^{2}}$
From equation (iv),

$∵ F_{net} = \sqrt{F_{7}^{2} + F_{8}^{2} + 2 F_{7} F_{8} \cos θ}$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2} + 2 \frac{GMm}{d^{2}} \times \frac{GMm}{d^{2}} \cos θ}$
$= \frac{GMm}{d^{2}} \sqrt{1 + 1 + 2 \cos θ}$
$= \frac{GMm}{d^{2}} (\sqrt{2 (1 + \cos θ)})$
$= \sqrt{2} \frac{GMm}{d^{2}} \sqrt{1 + \cos θ} = \sqrt{2} \frac{GMm}{d^{2}} \sqrt{2} \cos \frac{θ}{2}$
$= 2 \frac{GMm}{d^{2}} \cos \frac{θ}{2}$
For, $0 < θ < 90^{\circ}$ then,
$∴ F_{net} > \sqrt{2} \frac{GMm}{d^{2}}$
Thus, the rank of gravitational force from least to greatest, $2 < 1 < 3 < 4$

AMU-2018

Gravitation

138274 The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order.

1 (i), (iii) = (iv), (ii)

2 (i) = (iii), (ii), (iv)

3 (i), (ii), (iii), (iv)

4 (iv), (iii), (ii), (ii)

Explanation:

A We know that,
Gravitational force $(F) = \frac{{Gm}_{1} m_{2}}{r^{2}}$
From figure (i),
$F_{1} = \frac{Gmm}{d^{2}} + \frac{Gmm}{D^{2}} = {Gm}^{2} (\frac{1}{d^{2}} + \frac{1}{D^{2}})$
From figure (ii),
$F_{2} = \frac{Gmm}{d^{2}} - \frac{Gmm}{(D - d)^{2}} = {Gm}^{2} (\frac{1}{d^{2}} - \frac{1}{D^{2}})$
From figure (iii),
$F_{3} = \sqrt{{(\frac{G m m}{d^{2}})}^{2} + {(\frac{G m m}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} + {(\frac{1}{(D)^{2}})}^{2}}$
From figure (iv),
$F_{4} = \sqrt{{(\frac{Gmm}{d^{2}})}^{2} - {(\frac{Gmm}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} - {(\frac{1}{D^{2}})}^{2}}$
Hence, the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order-
$(i), (iii) = (iv), (ii)$

AMU-2013

Gravitation

138270 Suppose the force of gravitation between two bodies of equal masses is $F$ . If each mass is doubled keeping the distance of separation between them unchanged, the force would become

1

F

2

2 F

3

4 F

4

\frac{1}{4} F

Explanation:

C Two bodies are of equal mass.
So, $m_{1} = m_{2} = m$
$m_{1} =$ mass of $1^{st}$ body, $m_{2} =$ mass of $2^{nd}$ body and $r =$ distance between the two bodies
Then, $F = G \frac{m_{1} m_{2}}{r^{2}}$
$F = \frac{{Gm}^{2}}{r^{2}}$
Now masses are doubled and distance is the same
$m_{1} = m_{2} = 2 m$
The magnitude of the gravitational force $F$ using equation (i)
$F = G \frac{2 m \times 2 m}{r^{2}}$
$F^{1} = 4 \times \frac{{Gm}^{2}}{r^{2}} = 4 F$

NDA (II) 2019

Gravitation

138271 Which one of the following statements is true for the relation $F = \frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)

1 The quantity

G

depends on the local value of

g

, acceleration due to gravity

2 The quantity

G

is greatest at the surface of the Earth

3 The quantity G is used only when earth is one of the two masses

4 The quantity

G

is a universal constant

Explanation:

D Gravitational Force $(F) = G \frac{m_{1} m_{2}}{r^{2}}$
Where, $m_{1} =$ mass of first body
$m_{2} =$ mass of second body
$r =$ distance between two body
$G =$ Gravitational constant
$G$ is the universal gravitational constant which remain constant at all places in the universe $G$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart.
The value of $G = 6.67 \times 10^{- 11} {Nm}^{2} / {kg}^{2}$

NDA (I) 2017

Gravitation

138273 Three particles, two with masses $m$ and one with mass $M$ , might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$ , least to greatest.

1

1, 2, 3, 4

2

2, 1, 3, 4

3

2, 1, 4, 3

4 2, 3, 4, 1

Explanation:

B From figure (i),

$∵ F_{net} = F_{1} + F_{2}$
$= \frac{GMm}{d^{2}} + \frac{GMm}{(2 d)^{2}}$
$= \frac{GMm}{d^{2}} (1 + \frac{1}{4})$
$= \frac{5}{4} \frac{GMm}{d^{2}} = 1.25 \frac{GMm}{d^{2}}$
From figure (ii),

$∵ F_{net} = F_{4} - F_{3}$
$= \frac{GMm}{d^{2}} - \frac{GMm}{d^{2}} = 0$
From figure (iii),

$∵ F_{net} = \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos θ}$
$= \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos 90^{\circ}} (∵ \cos 90^{\circ} = 0)$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2}}$
$F_{net} = \sqrt{2} \frac{GMm}{d^{2}} = 1.414 \frac{GMm}{d^{2}}$
From equation (iv),

$∵ F_{net} = \sqrt{F_{7}^{2} + F_{8}^{2} + 2 F_{7} F_{8} \cos θ}$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2} + 2 \frac{GMm}{d^{2}} \times \frac{GMm}{d^{2}} \cos θ}$
$= \frac{GMm}{d^{2}} \sqrt{1 + 1 + 2 \cos θ}$
$= \frac{GMm}{d^{2}} (\sqrt{2 (1 + \cos θ)})$
$= \sqrt{2} \frac{GMm}{d^{2}} \sqrt{1 + \cos θ} = \sqrt{2} \frac{GMm}{d^{2}} \sqrt{2} \cos \frac{θ}{2}$
$= 2 \frac{GMm}{d^{2}} \cos \frac{θ}{2}$
For, $0 < θ < 90^{\circ}$ then,
$∴ F_{net} > \sqrt{2} \frac{GMm}{d^{2}}$
Thus, the rank of gravitational force from least to greatest, $2 < 1 < 3 < 4$

AMU-2018

Gravitation

138274 The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order.

1 (i), (iii) = (iv), (ii)

2 (i) = (iii), (ii), (iv)

3 (i), (ii), (iii), (iv)

4 (iv), (iii), (ii), (ii)

Explanation:

A We know that,
Gravitational force $(F) = \frac{{Gm}_{1} m_{2}}{r^{2}}$
From figure (i),
$F_{1} = \frac{Gmm}{d^{2}} + \frac{Gmm}{D^{2}} = {Gm}^{2} (\frac{1}{d^{2}} + \frac{1}{D^{2}})$
From figure (ii),
$F_{2} = \frac{Gmm}{d^{2}} - \frac{Gmm}{(D - d)^{2}} = {Gm}^{2} (\frac{1}{d^{2}} - \frac{1}{D^{2}})$
From figure (iii),
$F_{3} = \sqrt{{(\frac{G m m}{d^{2}})}^{2} + {(\frac{G m m}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} + {(\frac{1}{(D)^{2}})}^{2}}$
From figure (iv),
$F_{4} = \sqrt{{(\frac{Gmm}{d^{2}})}^{2} - {(\frac{Gmm}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} - {(\frac{1}{D^{2}})}^{2}}$
Hence, the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order-
$(i), (iii) = (iv), (ii)$

AMU-2013

Gravitation

138270 Suppose the force of gravitation between two bodies of equal masses is $F$ . If each mass is doubled keeping the distance of separation between them unchanged, the force would become

1

F

2

2 F

3

4 F

4

\frac{1}{4} F

Explanation:

C Two bodies are of equal mass.
So, $m_{1} = m_{2} = m$
$m_{1} =$ mass of $1^{st}$ body, $m_{2} =$ mass of $2^{nd}$ body and $r =$ distance between the two bodies
Then, $F = G \frac{m_{1} m_{2}}{r^{2}}$
$F = \frac{{Gm}^{2}}{r^{2}}$
Now masses are doubled and distance is the same
$m_{1} = m_{2} = 2 m$
The magnitude of the gravitational force $F$ using equation (i)
$F = G \frac{2 m \times 2 m}{r^{2}}$
$F^{1} = 4 \times \frac{{Gm}^{2}}{r^{2}} = 4 F$

NDA (II) 2019

Gravitation

138271 Which one of the following statements is true for the relation $F = \frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)

1 The quantity

G

depends on the local value of

g

, acceleration due to gravity

2 The quantity

G

is greatest at the surface of the Earth

3 The quantity G is used only when earth is one of the two masses

4 The quantity

G

is a universal constant

Explanation:

D Gravitational Force $(F) = G \frac{m_{1} m_{2}}{r^{2}}$
Where, $m_{1} =$ mass of first body
$m_{2} =$ mass of second body
$r =$ distance between two body
$G =$ Gravitational constant
$G$ is the universal gravitational constant which remain constant at all places in the universe $G$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart.
The value of $G = 6.67 \times 10^{- 11} {Nm}^{2} / {kg}^{2}$

NDA (I) 2017

Gravitation

138273 Three particles, two with masses $m$ and one with mass $M$ , might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$ , least to greatest.

1

1, 2, 3, 4

2

2, 1, 3, 4

3

2, 1, 4, 3

4 2, 3, 4, 1

Explanation:

B From figure (i),

$∵ F_{net} = F_{1} + F_{2}$
$= \frac{GMm}{d^{2}} + \frac{GMm}{(2 d)^{2}}$
$= \frac{GMm}{d^{2}} (1 + \frac{1}{4})$
$= \frac{5}{4} \frac{GMm}{d^{2}} = 1.25 \frac{GMm}{d^{2}}$
From figure (ii),

$∵ F_{net} = F_{4} - F_{3}$
$= \frac{GMm}{d^{2}} - \frac{GMm}{d^{2}} = 0$
From figure (iii),

$∵ F_{net} = \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos θ}$
$= \sqrt{F_{5}^{2} + F_{6}^{2} + 2 F_{5} F_{6} \cos 90^{\circ}} (∵ \cos 90^{\circ} = 0)$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2}}$
$F_{net} = \sqrt{2} \frac{GMm}{d^{2}} = 1.414 \frac{GMm}{d^{2}}$
From equation (iv),

$∵ F_{net} = \sqrt{F_{7}^{2} + F_{8}^{2} + 2 F_{7} F_{8} \cos θ}$
$= \sqrt{{(\frac{GMm}{d^{2}})}^{2} + {(\frac{GMm}{d^{2}})}^{2} + 2 \frac{GMm}{d^{2}} \times \frac{GMm}{d^{2}} \cos θ}$
$= \frac{GMm}{d^{2}} \sqrt{1 + 1 + 2 \cos θ}$
$= \frac{GMm}{d^{2}} (\sqrt{2 (1 + \cos θ)})$
$= \sqrt{2} \frac{GMm}{d^{2}} \sqrt{1 + \cos θ} = \sqrt{2} \frac{GMm}{d^{2}} \sqrt{2} \cos \frac{θ}{2}$
$= 2 \frac{GMm}{d^{2}} \cos \frac{θ}{2}$
For, $0 < θ < 90^{\circ}$ then,
$∴ F_{net} > \sqrt{2} \frac{GMm}{d^{2}}$
Thus, the rank of gravitational force from least to greatest, $2 < 1 < 3 < 4$

AMU-2018

Gravitation

138274 The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order.

1 (i), (iii) = (iv), (ii)

2 (i) = (iii), (ii), (iv)

3 (i), (ii), (iii), (iv)

4 (iv), (iii), (ii), (ii)

Explanation:

A We know that,
Gravitational force $(F) = \frac{{Gm}_{1} m_{2}}{r^{2}}$
From figure (i),
$F_{1} = \frac{Gmm}{d^{2}} + \frac{Gmm}{D^{2}} = {Gm}^{2} (\frac{1}{d^{2}} + \frac{1}{D^{2}})$
From figure (ii),
$F_{2} = \frac{Gmm}{d^{2}} - \frac{Gmm}{(D - d)^{2}} = {Gm}^{2} (\frac{1}{d^{2}} - \frac{1}{D^{2}})$
From figure (iii),
$F_{3} = \sqrt{{(\frac{G m m}{d^{2}})}^{2} + {(\frac{G m m}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} + {(\frac{1}{(D)^{2}})}^{2}}$
From figure (iv),
$F_{4} = \sqrt{{(\frac{Gmm}{d^{2}})}^{2} - {(\frac{Gmm}{D^{2}})}^{2}} = {Gm}^{2} \sqrt{{(\frac{1}{d^{2}})}^{2} - {(\frac{1}{D^{2}})}^{2}}$
Hence, the magnitude of the net gravitational force on the particle labeled $m$ , in decreasing order-
$(i), (iii) = (iv), (ii)$

AMU-2013

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