QBManager

PHXI08:GRAVITATION

359884 The gravitational field in a region is given by:
$\vec{E} = (5 N / k g) \hat{i} + (12 N / k g) \hat{j}$
If the potential at the origin is taken to be zero, Then the ratio of the potential at the points $(12 m, 0)$ and $(0.5 m)$ is :

1 Zero

2 1

3

\frac{144}{25}

4

\frac{25}{144}

Explanation:

From question,
$E_{x} = 5 N / k g$ and $E_{y} = 12 N / k g$
$V = E d$
$∴ V_{(12 m, 0)} = E_{x} \times 12 J / k g$
and $V_{(0, 5 m)} = E_{y} \times 5 J / k g$
(Given : potential at the origin is zero)
$∴ \frac{V_{(12 m, 0)}}{V_{(0, 2 m)}} = \frac{E_{x} \times 12}{E_{y} \times 5} = \frac{5 \times 12}{12 \times 5} = 1$

JEE - 2013

PHXI08:GRAVITATION

359885 In some region, the gravitational field is zero. Then gravitational potential at that region

1 Must be zero

2 Can not be zero

3 Must be constant

4 Continuously varies

Explanation:

Gravitational field is
$E = - \frac{d V}{d r}$
If $E =$ then $V$ is constant.

PHXI08:GRAVITATION

359886 If $W_{1}, W_{2}$ and $W_{3}$ represent the work done in moving a particle from $A$ and $B$ along three different paths 1, 2 and 3, respectively (As shown in the figure) in the gravitational field of a point mass $m$ , find the correct relation
between $W_{1}, W_{2}$ and $W_{3}$ .
supporting img

1

W_{1} > W_{2} > W_{3}

2

W_{1} = W_{2} = W_{3}

3

W_{1} < W_{2} < W_{3}

4

W_{2} > W_{1} > W_{3}

Explanation:

As gravitational force is a conservative force, work done is independent of path.
$W_{1} = W_{2} = W_{3}$

PHXI08:GRAVITATION

359887 If the gravitational field in the space is given as $(- \frac{K}{r^{2}})$ . Taking the reference point to be at $r = 2 c m$ with gravitational potential $V = 10 J / k g$ . Find the gravitational potential at $r = 3 c m$ in SI unit (Given, that $K = 6 J c m / k g$ )

1 11

2 10

3 12

4 9

Explanation:

We know, $g = (\frac{- K}{r^{2}})$
Relation between gravitational field and gravitational potential is given by
$\frac{- d V}{d r} = E_{g}$ or $g$
or $\frac{- d V}{d r} = \frac{- K}{r^{2}}$ or $\int_{V_{i}}^{V_{f}} d V = K \int_{2}^{3} \frac{1}{r^{2}} d r$
or $[V_{3} - V_{2}] = - K [\frac{1}{3} - \frac{1}{2}]$
or $V_{3} - 10 = - 6 [\frac{2 - 3}{6}]$
or $V_{3} = 1 + 10 = 11 V$

JEE - 2023

PHXI08:GRAVITATION

359884 The gravitational field in a region is given by:
$\vec{E} = (5 N / k g) \hat{i} + (12 N / k g) \hat{j}$
If the potential at the origin is taken to be zero, Then the ratio of the potential at the points $(12 m, 0)$ and $(0.5 m)$ is :

1 Zero

2 1

3

\frac{144}{25}

4

\frac{25}{144}

Explanation:

From question,
$E_{x} = 5 N / k g$ and $E_{y} = 12 N / k g$
$V = E d$
$∴ V_{(12 m, 0)} = E_{x} \times 12 J / k g$
and $V_{(0, 5 m)} = E_{y} \times 5 J / k g$
(Given : potential at the origin is zero)
$∴ \frac{V_{(12 m, 0)}}{V_{(0, 2 m)}} = \frac{E_{x} \times 12}{E_{y} \times 5} = \frac{5 \times 12}{12 \times 5} = 1$

JEE - 2013

PHXI08:GRAVITATION

359885 In some region, the gravitational field is zero. Then gravitational potential at that region

1 Must be zero

2 Can not be zero

3 Must be constant

4 Continuously varies

Explanation:

Gravitational field is
$E = - \frac{d V}{d r}$
If $E =$ then $V$ is constant.

PHXI08:GRAVITATION

359886 If $W_{1}, W_{2}$ and $W_{3}$ represent the work done in moving a particle from $A$ and $B$ along three different paths 1, 2 and 3, respectively (As shown in the figure) in the gravitational field of a point mass $m$ , find the correct relation
between $W_{1}, W_{2}$ and $W_{3}$ .
supporting img

1

W_{1} > W_{2} > W_{3}

2

W_{1} = W_{2} = W_{3}

3

W_{1} < W_{2} < W_{3}

4

W_{2} > W_{1} > W_{3}

Explanation:

As gravitational force is a conservative force, work done is independent of path.
$W_{1} = W_{2} = W_{3}$

PHXI08:GRAVITATION

359887 If the gravitational field in the space is given as $(- \frac{K}{r^{2}})$ . Taking the reference point to be at $r = 2 c m$ with gravitational potential $V = 10 J / k g$ . Find the gravitational potential at $r = 3 c m$ in SI unit (Given, that $K = 6 J c m / k g$ )

1 11

2 10

3 12

4 9

Explanation:

We know, $g = (\frac{- K}{r^{2}})$
Relation between gravitational field and gravitational potential is given by
$\frac{- d V}{d r} = E_{g}$ or $g$
or $\frac{- d V}{d r} = \frac{- K}{r^{2}}$ or $\int_{V_{i}}^{V_{f}} d V = K \int_{2}^{3} \frac{1}{r^{2}} d r$
or $[V_{3} - V_{2}] = - K [\frac{1}{3} - \frac{1}{2}]$
or $V_{3} - 10 = - 6 [\frac{2 - 3}{6}]$
or $V_{3} = 1 + 10 = 11 V$

JEE - 2023

PHXI08:GRAVITATION

359884 The gravitational field in a region is given by:
$\vec{E} = (5 N / k g) \hat{i} + (12 N / k g) \hat{j}$
If the potential at the origin is taken to be zero, Then the ratio of the potential at the points $(12 m, 0)$ and $(0.5 m)$ is :

1 Zero

2 1

3

\frac{144}{25}

4

\frac{25}{144}

Explanation:

From question,
$E_{x} = 5 N / k g$ and $E_{y} = 12 N / k g$
$V = E d$
$∴ V_{(12 m, 0)} = E_{x} \times 12 J / k g$
and $V_{(0, 5 m)} = E_{y} \times 5 J / k g$
(Given : potential at the origin is zero)
$∴ \frac{V_{(12 m, 0)}}{V_{(0, 2 m)}} = \frac{E_{x} \times 12}{E_{y} \times 5} = \frac{5 \times 12}{12 \times 5} = 1$

JEE - 2013

PHXI08:GRAVITATION

359885 In some region, the gravitational field is zero. Then gravitational potential at that region

1 Must be zero

2 Can not be zero

3 Must be constant

4 Continuously varies

Explanation:

Gravitational field is
$E = - \frac{d V}{d r}$
If $E =$ then $V$ is constant.

PHXI08:GRAVITATION

359886 If $W_{1}, W_{2}$ and $W_{3}$ represent the work done in moving a particle from $A$ and $B$ along three different paths 1, 2 and 3, respectively (As shown in the figure) in the gravitational field of a point mass $m$ , find the correct relation
between $W_{1}, W_{2}$ and $W_{3}$ .
supporting img

1

W_{1} > W_{2} > W_{3}

2

W_{1} = W_{2} = W_{3}

3

W_{1} < W_{2} < W_{3}

4

W_{2} > W_{1} > W_{3}

Explanation:

As gravitational force is a conservative force, work done is independent of path.
$W_{1} = W_{2} = W_{3}$

PHXI08:GRAVITATION

359887 If the gravitational field in the space is given as $(- \frac{K}{r^{2}})$ . Taking the reference point to be at $r = 2 c m$ with gravitational potential $V = 10 J / k g$ . Find the gravitational potential at $r = 3 c m$ in SI unit (Given, that $K = 6 J c m / k g$ )

1 11

2 10

3 12

4 9

Explanation:

We know, $g = (\frac{- K}{r^{2}})$
Relation between gravitational field and gravitational potential is given by
$\frac{- d V}{d r} = E_{g}$ or $g$
or $\frac{- d V}{d r} = \frac{- K}{r^{2}}$ or $\int_{V_{i}}^{V_{f}} d V = K \int_{2}^{3} \frac{1}{r^{2}} d r$
or $[V_{3} - V_{2}] = - K [\frac{1}{3} - \frac{1}{2}]$
or $V_{3} - 10 = - 6 [\frac{2 - 3}{6}]$
or $V_{3} = 1 + 10 = 11 V$

JEE - 2023

PHXI08:GRAVITATION

359884 The gravitational field in a region is given by:
$\vec{E} = (5 N / k g) \hat{i} + (12 N / k g) \hat{j}$
If the potential at the origin is taken to be zero, Then the ratio of the potential at the points $(12 m, 0)$ and $(0.5 m)$ is :

1 Zero

2 1

3

\frac{144}{25}

4

\frac{25}{144}

Explanation:

From question,
$E_{x} = 5 N / k g$ and $E_{y} = 12 N / k g$
$V = E d$
$∴ V_{(12 m, 0)} = E_{x} \times 12 J / k g$
and $V_{(0, 5 m)} = E_{y} \times 5 J / k g$
(Given : potential at the origin is zero)
$∴ \frac{V_{(12 m, 0)}}{V_{(0, 2 m)}} = \frac{E_{x} \times 12}{E_{y} \times 5} = \frac{5 \times 12}{12 \times 5} = 1$

JEE - 2013

PHXI08:GRAVITATION

359885 In some region, the gravitational field is zero. Then gravitational potential at that region

1 Must be zero

2 Can not be zero

3 Must be constant

4 Continuously varies

Explanation:

Gravitational field is
$E = - \frac{d V}{d r}$
If $E =$ then $V$ is constant.

PHXI08:GRAVITATION

359886 If $W_{1}, W_{2}$ and $W_{3}$ represent the work done in moving a particle from $A$ and $B$ along three different paths 1, 2 and 3, respectively (As shown in the figure) in the gravitational field of a point mass $m$ , find the correct relation
between $W_{1}, W_{2}$ and $W_{3}$ .
supporting img

1

W_{1} > W_{2} > W_{3}

2

W_{1} = W_{2} = W_{3}

3

W_{1} < W_{2} < W_{3}

4

W_{2} > W_{1} > W_{3}

Explanation:

As gravitational force is a conservative force, work done is independent of path.
$W_{1} = W_{2} = W_{3}$

PHXI08:GRAVITATION

359887 If the gravitational field in the space is given as $(- \frac{K}{r^{2}})$ . Taking the reference point to be at $r = 2 c m$ with gravitational potential $V = 10 J / k g$ . Find the gravitational potential at $r = 3 c m$ in SI unit (Given, that $K = 6 J c m / k g$ )

1 11

2 10

3 12

4 9

Explanation:

We know, $g = (\frac{- K}{r^{2}})$
Relation between gravitational field and gravitational potential is given by
$\frac{- d V}{d r} = E_{g}$ or $g$
or $\frac{- d V}{d r} = \frac{- K}{r^{2}}$ or $\int_{V_{i}}^{V_{f}} d V = K \int_{2}^{3} \frac{1}{r^{2}} d r$
or $[V_{3} - V_{2}] = - K [\frac{1}{3} - \frac{1}{2}]$
or $V_{3} - 10 = - 6 [\frac{2 - 3}{6}]$
or $V_{3} = 1 + 10 = 11 V$

JEE - 2023