Gravitation Test

Result

Gravitation Test - 31

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The escape velocity for a planet is $$v_e$$. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be

A
$$v_a$$

B
$$1.5 v_e$$

C
$$\sqrt{1.5} v_e$$

D
$$2 v_e$$

Solution

Within the earth's body the gravitational potential at distance $$r$$ from the center is given as $$V=2\pi G\rho(r^2-3R^2)/3$$.
where $$\rho$$ is the density of earth, $$R$$ is the earth's radius.
At the center of earth, $$r=0$$.
So the potential is $$V_c=-2\pi G\rho\times3R^2/3=-3GM/2R$$.
where $$M=\rho\times4\pi R^3/3$$
At very large distance, potential $$V_i=0$$
So gain in kinetic energy = change in potential energy.
$$\Rightarrow mv^2/2=0-(-3GM/2R)=3GM/2R=1.5GM/R\Rightarrow v=\sqrt{1.5}v_e$$
where $$v_e=\sqrt{GM/R}$$ is the escape velocity.
Question 2
1 / -0

Find the height at which the weight will be same as at the same depth from the surface of the earth.

A
$$\displaystyle \frac{R}{2}$$

B
$$\sqrt{5} R - R$$

C
$$\displaystyle \frac{\sqrt{5} R - R}{2}$$

D
$$\displaystyle \frac{\sqrt{3}R - R}{2}$$

Solution

Let the height and depth be 'd'
$$\Rightarrow \quad \dfrac { g }{ { \left( 1+\dfrac { d }{ R } \right) }^{ 2 } } =g\left( 1-\dfrac { d }{ R } \right) $$
$$\Rightarrow { \left( 1+\dfrac { d }{ R } \right) }^{ 2 }\left( 1-\dfrac { d }{ R } \right) =1$$
$$\Rightarrow { \left( 1+\dfrac { { d }^{ 2 } }{ { R }^{ 2 } } +\dfrac { 2d }{ R } \right) }\left( 1-\dfrac { d }{ R } \right) =1\\ \Rightarrow { \left( \dfrac { d }{ R } \right) }^{ 3 }+{ \left( \dfrac { d }{ R } \right) }^{ 2 }-{ \left( \dfrac { d }{ R } \right) }=0\\ \Rightarrow \quad d=\dfrac { \sqrt { 5 } R-R }{ 2 } $$
Question 3
1 / -0

Use the assumptions of the previous question. An object weighed by a spring balance at the equator gives the same reading as a reading taken at a depth d below the earth's surface at a pole $$(d << R)$$. The value of $$d$$ is

A
$$\displaystyle \dfrac{\omega^2 R^2}{ge}$$

B
$$\displaystyle \dfrac{\omega^2 R^2}{2g}$$

C
$$\displaystyle \dfrac{2\omega^2 R^2}{g}$$

D
$$\displaystyle \dfrac{\sqrt{Rg}}{\omega}$$

Solution

Let $$g_e$$ be the acceleration due to gravity on surface of earth, $$g'$$ be the acceleration due to gravity at depth d below the surface of earth and is given by
$$g' = g_e \left(1 - \dfrac{d}{R} \right)$$ ................................(1)
The acceleration due to gravity of an object on below the surface of earth is
$$g' = g_e - \omega^2 R \cos \phi$$
where, $$\phi$$ is the solid angle at the center of earth sustained by the object.
At the equator, $$\phi = 0^\circ$$
$$\therefore \cos \phi = 1$$
Hence,
$$g' = g_e - \omega^2 R$$ ....................................(2)
From (1) and (2), we can write,
$$ g_e \left(1 - \dfrac{d}{R} \right) = g_e - \omega^2 R$$
$$ \Rightarrow \dfrac{dg_e}{R} = \omega^2 R$$
$$ \Rightarrow d = \dfrac{\omega^2 R}{g_e}$$
Question 4
1 / -0

The value of acceleration due to gravity at height $$h$$ from earth surface will become half its value on the surface if (R $$=$$ radius of earth)

A
$$h = R$$

B
$$h = 2R$$

C
$$h = (\sqrt{2} - 1)R$$

D
$$h = (\sqrt{2} + 1)R$$

Solution

The gravitational force acting on mass m on the surface of Earth is
$$F = \dfrac{GMm}{R^2}$$

$$mg = \dfrac{GMm}{R^2}$$

$$g = \dfrac{GM}{R^2}$$

$$gR^2 = GM$$.............$$(1)$$
where, $$g$$ is acceleration due to gravity, $$G$$ is gravitational constant, $$M$$ is mass of Earth and R is radius of Earth.
Now, value of g at height h is
$$g = \dfrac{GM}{(R+h)^2}$$

As per the problem,
$$\dfrac{g}{2} = \dfrac{GM}{(R+h)^2}$$

$$g(R+h)^2 = 2 gR^2$$................from $$(1)$$
$$(R+h)^2 = 2 R^2$$
Taking roots from both sides, we get
$$R+h = \sqrt 2 R$$
$$\therefore h = \sqrt 2 R - R$$
$$\therefore h = (\sqrt 2 - 1)R$$
Question 5
1 / -0

Consider a planet moving in an elliptical orbit round the sun. The work done on the planet by the gravitational force of the sun

A
is zero in any small part of the orbit.

B
is zero in some parts of the orbit.

C
is zero in one complete revolution.

D
is zero in no part of the motion.

Solution

The work done is zero at points A and B, since the force acts along the AB axis and the motion is perpendicular to it($$W=|F||S| \cos \theta$$)
Question 6
1 / -0

The gravitational potential difference between the surface of a planet and a point $$20 m$$ above the surface is $$2 Joule/Kg$$. If the gravitational field is uniform then the work done in carrying a $$5 Kg$$ body to a height of $$4 m$$ above the surface is

A
$$2 Joule$$

B
$$20 Joule$$

C
$$40 Joule$$

D
$$10 Joule$$

Solution

since it is given that gravitational field is uniform
gradient of potential$$=\dfrac { Potential\ diff. }{ Dist.\ between\ points }=\dfrac { 2 }{ 20 } =0.1\ J/Kg s$$
Total work done $$=mass\times distance\times pot.\ gradient=5\times 4\times 0.1=2\ Joule$$
Question 7
1 / -0

The magnitude of the potential energy per unit mass of an object at the surface of the earth is given $$U$$. Then, the escape velocity for the object is given by

A
$$v_e = 2U$$

B
$$v_e = \sqrt{U}$$

C
$$v_e = \sqrt{2U}$$

D
$$v_e = (2U)^2$$

Solution

Potential energy of unit mass of body at earth surface U = $$\dfrac{-GM(1)}{R} = \dfrac{-GM}{R}$$
escape velocity $$v_e = (\dfrac{2GM}{R})^{0.5} = (2U)^{0.5}$$
Question 8
1 / -0

The value of G for two bodies in vacuum is $$6.67 \times 10^{-11} N-m^2/Kg^2$$. Its value in a dense medium of density $$10^{10} gm/cm^3$$ will be:

A
$$6.67 \times 10^{-11} N-m^2 /Kg$$

B
$$6.67 \times 10^{-31} N-m^2 /Kg$$

C
$$6.67 \times 10^{-21} N-m^2 /Kg$$

D
$$6.67 \times 10^{-10} N-m^2 /Kg$$

Solution

G is a Universal Gravitational constant and is same every where in the universe. Thus,
G $$ = 6.67 \times 10^{-11} N.m^{2}.kg^{-2}$$
Question 9
1 / -0

The potential energy of a rocket of mass $$100\ kg$$ at height $$10^7\ m$$ from earth surface is $$4 \times 10^9\ Joule$$. The weight of the rocket at height $$10^9$$ will be

A
$$4 \times 10^{-2} N$$

B
$$4 \times 10^{-3} N$$

C
$$8 \times 10^{-2} N$$

D
$$8 \times 10^{-3} N$$

Solution

Distance $$R = r + h = 0.64 \times 10^{7} + 10^{7} = 1.64 \times 10^7 m$$
Now$$,\ U = \dfrac{GMm}{R} = \dfrac{GM \times 100}{1.64 \times 10^7} = GM \times .609 \times 10^{-5}$$

But, given $$U = 4 \times 10^9 \Rightarrow GM \times 0.609 \times 10^{-5} = 4 \times 10^9$$
$$GM = 6.57 \times 10^{14}$$
now at height $$10^9 m$$, $$R = r+h = 100.64 \times 10^7$$
weight $$mg = \dfrac{GMm}{R^2} = \dfrac{6.57 \times 10^{14} }{(100.64 \times 10^7)^{2}} = 4 \times 10^{-2}$$ N.
Question 10
1 / -0

A tunnel is dug along a diameter of earth. The force on a particle of mass $$m$$ and distance $$x$$ from the centre in this tunnel will be :

A
$$\displaystyle \frac{GM_e m}{R^3 x}$$

B
$$\displaystyle \frac{GM_e mR^3}{x}$$

C
$$\displaystyle \frac{GM_e mx}{R^2}$$

D
$$\displaystyle \frac{GM_e mx}{R^3}$$

Solution

Field at a distance $$x$$ from the center will be equal to field at a distance $$R-x$$ from the earth's surface.

So
$$g_x=g(1-\displaystyle\frac{(R-x)}{R})=\frac{gx}{R}$$

or

$$F=\displaystyle\frac{mgx}{R}=\frac{mx}{R}\frac{GM_e}{R^2}=\frac{GM_e}{R^3}mx$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Gravitation Test - 31

Result

Gravitation Test - 31

Score

-

Rank

-

SHARING IS CARING

Question 1

$$v_a$$

$$1.5 v_e$$

$$\sqrt{1.5} v_e$$

$$2 v_e$$

Solution

Question 2

$$\displaystyle \frac{R}{2}$$

$$\sqrt{5} R - R$$

$$\displaystyle \frac{\sqrt{5} R - R}{2}$$

$$\displaystyle \frac{\sqrt{3}R - R}{2}$$

Solution

Question 3

$$\displaystyle \dfrac{\omega^2 R^2}{ge}$$

$$\displaystyle \dfrac{\omega^2 R^2}{2g}$$

$$\displaystyle \dfrac{2\omega^2 R^2}{g}$$

$$\displaystyle \dfrac{\sqrt{Rg}}{\omega}$$

Solution

Question 4

$$h = R$$

$$h = 2R$$

$$h = (\sqrt{2} - 1)R$$

$$h = (\sqrt{2} + 1)R$$

Solution

Question 5

is zero in any small part of the orbit.

is zero in some parts of the orbit.

is zero in one complete revolution.

is zero in no part of the motion.

Solution

Question 6

$$2 Joule$$

$$20 Joule$$

$$40 Joule$$

$$10 Joule$$

Solution

Question 7

$$v_e = 2U$$

$$v_e = \sqrt{U}$$

$$v_e = \sqrt{2U}$$

$$v_e = (2U)^2$$

Solution

Question 8

$$6.67 \times 10^{-11} N-m^2 /Kg$$

$$6.67 \times 10^{-31} N-m^2 /Kg$$

$$6.67 \times 10^{-21} N-m^2 /Kg$$

$$6.67 \times 10^{-10} N-m^2 /Kg$$

Solution

Question 9

$$4 \times 10^{-2} N$$

$$4 \times 10^{-3} N$$

$$8 \times 10^{-2} N$$

$$8 \times 10^{-3} N$$

Solution

Question 10

$$\displaystyle \frac{GM_e m}{R^3 x}$$

$$\displaystyle \frac{GM_e mR^3}{x}$$

$$\displaystyle \frac{GM_e mx}{R^2}$$

$$\displaystyle \frac{GM_e mx}{R^3}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.