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Gravitation

270546 A $3 kg$ mass and a $4 kg$ mass are placed on $x$ and $y$ axes at a distance of 1 metre from the origin and a $1 kg$ mass is placed at the origin. Then the resultant gravitational force on $1 kg$ mass is

1

7 G

2 G

3

5 G

4

3 G

Explanation:

$θ = 90^{\circ}, F_{1} = \frac{3 G}{1^{2}}, F_{2} = \frac{4 G}{1^{2}}$ $F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} Cos θ}$

Gravitation

270592 Gravitational force between two point masses $m$ and $M$ separated by a distance $r$ is $F$ . Now if a point mass $3 m$ is placed next to $m$ , the total force on $M$ will be

1

F

2

2 F

3

3 F

4

4 F

Explanation:

$F = \frac{G M m}{r^{2}}; F^{'} = \frac{G M (m + 2 m)}{r^{2}}$

Gravitation

270593 If three particles,each of mass $M$ are placed at the three corners of an equilateral triangle of side $a$ , the force exerted by this system on another particle of mass $M$ placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.

1

0, \frac{4 G M^{2}}{3 a^{2}}

2

\frac{4 G M^{2}}{3 a^{2}}, 0

3

\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}

4 0,0

Explanation:

Find individual forces and calculate resultant
$Use F = \frac{G m_{1} m_{2}}{R^{2}}$

Gravitation

270594 Two masses ' $M$ ' and ' $4 M$ ' are at a distance ' $r$ ' apart on the line joining them, ' $P$ ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' $P$ ' from the mass ' $M$ ' is

1

\frac{r}{5}

2

\frac{r}{3}

3

\frac{2 r}{3}

4

\frac{4 r}{5}

Explanation:

$x = \frac{d}{\sqrt{\frac{m_{2}}{m_{1}}} + 1}$

Gravitation

270546 A $3 kg$ mass and a $4 kg$ mass are placed on $x$ and $y$ axes at a distance of 1 metre from the origin and a $1 kg$ mass is placed at the origin. Then the resultant gravitational force on $1 kg$ mass is

1

7 G

2 G

3

5 G

4

3 G

Explanation:

$θ = 90^{\circ}, F_{1} = \frac{3 G}{1^{2}}, F_{2} = \frac{4 G}{1^{2}}$ $F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} Cos θ}$

Gravitation

270592 Gravitational force between two point masses $m$ and $M$ separated by a distance $r$ is $F$ . Now if a point mass $3 m$ is placed next to $m$ , the total force on $M$ will be

1

F

2

2 F

3

3 F

4

4 F

Explanation:

$F = \frac{G M m}{r^{2}}; F^{'} = \frac{G M (m + 2 m)}{r^{2}}$

Gravitation

270593 If three particles,each of mass $M$ are placed at the three corners of an equilateral triangle of side $a$ , the force exerted by this system on another particle of mass $M$ placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.

1

0, \frac{4 G M^{2}}{3 a^{2}}

2

\frac{4 G M^{2}}{3 a^{2}}, 0

3

\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}

4 0,0

Explanation:

Find individual forces and calculate resultant
$Use F = \frac{G m_{1} m_{2}}{R^{2}}$

Gravitation

270594 Two masses ' $M$ ' and ' $4 M$ ' are at a distance ' $r$ ' apart on the line joining them, ' $P$ ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' $P$ ' from the mass ' $M$ ' is

1

\frac{r}{5}

2

\frac{r}{3}

3

\frac{2 r}{3}

4

\frac{4 r}{5}

Explanation:

$x = \frac{d}{\sqrt{\frac{m_{2}}{m_{1}}} + 1}$

Gravitation

270546 A $3 kg$ mass and a $4 kg$ mass are placed on $x$ and $y$ axes at a distance of 1 metre from the origin and a $1 kg$ mass is placed at the origin. Then the resultant gravitational force on $1 kg$ mass is

1

7 G

2 G

3

5 G

4

3 G

Explanation:

$θ = 90^{\circ}, F_{1} = \frac{3 G}{1^{2}}, F_{2} = \frac{4 G}{1^{2}}$ $F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} Cos θ}$

Gravitation

270592 Gravitational force between two point masses $m$ and $M$ separated by a distance $r$ is $F$ . Now if a point mass $3 m$ is placed next to $m$ , the total force on $M$ will be

1

F

2

2 F

3

3 F

4

4 F

Explanation:

$F = \frac{G M m}{r^{2}}; F^{'} = \frac{G M (m + 2 m)}{r^{2}}$

Gravitation

270593 If three particles,each of mass $M$ are placed at the three corners of an equilateral triangle of side $a$ , the force exerted by this system on another particle of mass $M$ placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.

1

0, \frac{4 G M^{2}}{3 a^{2}}

2

\frac{4 G M^{2}}{3 a^{2}}, 0

3

\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}

4 0,0

Explanation:

Find individual forces and calculate resultant
$Use F = \frac{G m_{1} m_{2}}{R^{2}}$

Gravitation

270594 Two masses ' $M$ ' and ' $4 M$ ' are at a distance ' $r$ ' apart on the line joining them, ' $P$ ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' $P$ ' from the mass ' $M$ ' is

1

\frac{r}{5}

2

\frac{r}{3}

3

\frac{2 r}{3}

4

\frac{4 r}{5}

Explanation:

$x = \frac{d}{\sqrt{\frac{m_{2}}{m_{1}}} + 1}$

Gravitation

270546 A $3 kg$ mass and a $4 kg$ mass are placed on $x$ and $y$ axes at a distance of 1 metre from the origin and a $1 kg$ mass is placed at the origin. Then the resultant gravitational force on $1 kg$ mass is

1

7 G

2 G

3

5 G

4

3 G

Explanation:

$θ = 90^{\circ}, F_{1} = \frac{3 G}{1^{2}}, F_{2} = \frac{4 G}{1^{2}}$ $F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} Cos θ}$

Gravitation

270592 Gravitational force between two point masses $m$ and $M$ separated by a distance $r$ is $F$ . Now if a point mass $3 m$ is placed next to $m$ , the total force on $M$ will be

1

F

2

2 F

3

3 F

4

4 F

Explanation:

$F = \frac{G M m}{r^{2}}; F^{'} = \frac{G M (m + 2 m)}{r^{2}}$

Gravitation

270593 If three particles,each of mass $M$ are placed at the three corners of an equilateral triangle of side $a$ , the force exerted by this system on another particle of mass $M$ placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.

1

0, \frac{4 G M^{2}}{3 a^{2}}

2

\frac{4 G M^{2}}{3 a^{2}}, 0

3

\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}

4 0,0

Explanation:

Find individual forces and calculate resultant
$Use F = \frac{G m_{1} m_{2}}{R^{2}}$

Gravitation

270594 Two masses ' $M$ ' and ' $4 M$ ' are at a distance ' $r$ ' apart on the line joining them, ' $P$ ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' $P$ ' from the mass ' $M$ ' is

1

\frac{r}{5}

2

\frac{r}{3}

3

\frac{2 r}{3}

4

\frac{4 r}{5}

Explanation:

$x = \frac{d}{\sqrt{\frac{m_{2}}{m_{1}}} + 1}$

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