Gravitation Test

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Gravitation Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The angular velocity of rotation of a star (mass M and radius R), such that the matter will start escaping from its equator is:

A
$$\sqrt{\dfrac{2GR}{M}}$$

B
$$\sqrt{\dfrac{2GM}{R^{3}}}$$

C
$$\sqrt{\dfrac{2GM}{R}}$$

D
$$\sqrt{\dfrac{2GM^{2}}{R}}$$

Solution

$$v_{esc} = \sqrt{ \dfrac{2GM}{R}}$$
$$ \omega R = \sqrt{ \dfrac{2GM}{R}}$$
$$ \therefore \omega = \sqrt{ \dfrac{2GM}{R^3}}$$
$$\omega =\displaystyle \dfrac{1}{R}\sqrt{\dfrac{2GM}{R}}$$
Question 2
1 / -0

A body of mass $$m$$ is raised from the surface of the earth to a height $$nR$$ ($$R$$-radius of earth). Magnitude of the change in the gravitational potential energy of the body is ($$g$$- acceleration due to gravity on the surface of earth) :

A
$$\left ( \dfrac{n}{n+1} \right )mgR$$

B
$$\left ( \dfrac{n-1}{n} \right )mgR$$

C
$$\dfrac{mgR}{n}$$

D
$$\dfrac{mgR}{\left ( n-1 \right )}$$

Solution

Change in P.E.$$ = \dfrac {-GMm}{(n+1)R} - (\dfrac{-GMm}{R}) = (\dfrac {GMm}{(n+1)R}) ( -1 + (n+1)) $$
$$= (\dfrac {GMm}{R}) \dfrac{(n)}{n+1}$$
But $$ g = GM/R^2$$ or $$GM = g R^2$$
Putting this value, we get Change in P.E. $$= mgR (\dfrac {n}{n+1})$$
Question 3
1 / -0

The work done to increase the radius of orbit of a satellite of mass $$m$$ revolving around a planet of mass $$M$$ from orbit of radius $$R$$ into another orbit of radius $$3R$$ is :

A
$$\dfrac{2GMm}{3R}$$

B
$$\dfrac{GMm}{R}$$

C
$$\dfrac{GMm}{6R}$$

D
$$\dfrac{GMm}{24R}$$

Solution

The gravitational force '$$F$$' on satellite of mass 'm' revolving around the planet in the orbit of radius '$$R$$' is
$$F = - \dfrac{GMm}{R^2} $$
Where
$$G$$ is gravitational constant
$$M$$ is Mass of the planet.
The work done in increasing the radius of orbit of the satellite through '$$dr$$' is
$$W = F dr $$
but work done is equal to change in gravitational potential '$$-dV$$'
therefore
$$ -dV = F dr $$
$$V = - \int_{R}^{3R}F dr$$
$$V = -\dfrac{GMm}{R^2}\int_{R}^{3R}-(F dr)$$
$$V = GMm\int_{R}^{3R}\dfrac {1}{R^2}dr$$
$$V = \dfrac{GMm}{R} - \dfrac{GMm}{3R}$$
$$V = \dfrac{2GMm}{3R}$$
Thus work done to increase the radius of orbit of satellite is $$ \dfrac{2GMm}{3R}$$
Question 4
1 / -0

If $$g$$ is acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is:

A
$$2 mgR$$

B
$$mgR$$

C
$$\dfrac{mgR}{4}$$

D
$$\dfrac{mgR}{2}$$

Solution

Value of g is given by: $$g=\dfrac { { GM }_{ e } }{ { R }^{ 2 } } $$
Initial Potential Energy: $${ U }_{ i }=-\dfrac { { GM }_{ e }m }{ { R } } =-MgR$$
Final Potential Energy: $$U_f =-\dfrac { { GM }_{ e }m }{ { 2R } } =-MgR/2$$
Therefore Gain in Potential Energy: $${ U }_{ f }-{ U }_{ i }=-MgR/2-(-MgR)=\dfrac{MgR}{2}$$
Question 5
1 / -0

Energy required to move a body of mass m from an orbit of radius 2R to 3R is:

A
$$\dfrac{GMm}{12R^{2}}$$

B
$$\dfrac{GMm}{3R^{2}}$$

C
$$\dfrac{GMm}{8R}$$

D
$$\dfrac{GMm}{6R}$$

Solution

$$\text{Energy required} =\text{(Final potential-Initial potential)m}$$
$$\text{Energy required}=-\dfrac{GMm}{3R}-\dfrac{-GMm}{2R}=\dfrac{GMm}{6R}$$
Question 6
1 / -0

The change in the P.E., when a body of mass $$m$$ is displaced from Earth's surface to a vertical height equal to radius of earth (g $$=$$ acceleration due to gravity on earth surface) is:

A
$$\dfrac{mgR}{2}$$

B
$$\dfrac{2mgR}{3}$$

C
$$\dfrac{3mgR}{4}$$

D
$$\dfrac{mgR}{3}$$

Solution

$$\displaystyle \Delta PE=\frac{GMm}{R}\left \{ 1-\frac{1}{2} \right \}=
\displaystyle \frac{GMm}{2R}$$
$$gR=\displaystyle \frac{GM}{R}$$
$$ \displaystyle \Delta PE=\frac{mgR}{2}$$
Question 7
1 / -0

If $$g$$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is

A
$$2mgR$$

B
$$\dfrac{1}{2}mgR$$

C
$$\dfrac{1}{4}mgR$$

D
$$mgR$$

Solution

Gravitational potential energy on the earth surface, $$U_{r}=\displaystyle \dfrac{-GMm}{R}$$

Gravitational potential energy at a height h above the earth's surface, $$U_{h}=\displaystyle \dfrac{-GMm}{R+h}$$

$$ U_{h}=\displaystyle \dfrac{-GMm}{R+R}=\dfrac{-GMm}{2R}$$

Gain in gravitational potential energy $$=U_{h}-U_{r}$$

$$=\displaystyle \dfrac{-GMm}{2R}-(\dfrac{-GMm}{R})=\dfrac{GMm}{R}-\dfrac{GMm}{2R}$$

$$=\displaystyle \dfrac{GMm}{2R}=\dfrac{1}{2}mgR$$
Question 8
1 / -0

A small body is initially at a distance $$r$$ from the centre of earth. $$r$$ is greater than the radius of the earth. If it takes $$W$$ joules of work to move the body from this position to another position at a distance $$2r$$ measured from the centre of earth, then how many joules would be required to move it from this position to a new position at a distance of $$3r$$ from the centre of the earth.

A
$$\dfrac{W}{5}$$

B
$$\dfrac{W}{3}$$

C
$$\dfrac{W}{2}$$

D
$$\dfrac{W}{6}$$

Solution

Work done to move the body in the gravitational field is equal to change in the potential energy of the body.
i.e.
$$W=-\dfrac {GMm}{2r}-\left(-\dfrac {GMm}{r}\right)$$

$$\Rightarrow W = \dfrac {GMm}{2r}$$
Now when the body is moved from $$2r$$ to $$3r$$, change in PE is
$$W'=-\dfrac {GMm}{3r}-\left(-\dfrac {GMm}{2r}\right)=\dfrac {1}{3}\dfrac {GMm}{2r}=\dfrac {W}{3}$$
Question 9
1 / -0

A satellite of mass $$m$$ moves along an elliptical path around the earth. The areal velocity of the satellite is proportional to

A
$$m$$

B
$$m^{-1}$$

C
$$m^{0}$$

D
$$m^{\frac{1}{2}}$$

Solution

The areal velocity $$ \dfrac{dA}{dt}$$ $$=$$constant
$$\dfrac{dA}{dt}=\dfrac{L}{2 m}$$

$$L=mvr \sin \theta$$ (angular momentum)

$$\dfrac{dA}{dt}=\dfrac{mvr \sin \theta}{2m}$$$$= v r \sin \theta$$

$$\therefore \dfrac{dA}{dt} \ \alpha\ m^0$$
Question 10
1 / -0

A person bring a mass of $$1 kg$$ from infinite to point $$A$$. Initially the mass was at rest but is moves a speed of $$2 m /s$$ as it reaches to $$A$$. The workdone by the person on mass is $$-3 J$$ the gravitational potential at $$A$$ is

A
$$-3 J / kg$$

B
$$-2 J / kg$$

C
$$-5 J / kg$$

D
$$-7 J / kg$$

Solution

$$K.E_{f}=1.2^2/2=2$$
Work Energy Theorem,
$$E_{f}-E_{i}=W$$
$$2+P.E-0=-3$$
$$P.E=mV=-5$$
$$V=-5$$ since $$m=1$$

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 26

Result

Gravitation Test - 26

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SHARING IS CARING

Question 1

$$\sqrt{\dfrac{2GR}{M}}$$

$$\sqrt{\dfrac{2GM}{R^{3}}}$$

$$\sqrt{\dfrac{2GM}{R}}$$

$$\sqrt{\dfrac{2GM^{2}}{R}}$$

Solution

Question 2

$$\left ( \dfrac{n}{n+1} \right )mgR$$

$$\left ( \dfrac{n-1}{n} \right )mgR$$

$$\dfrac{mgR}{n}$$

$$\dfrac{mgR}{\left ( n-1 \right )}$$

Solution

Question 3

$$\dfrac{2GMm}{3R}$$

$$\dfrac{GMm}{R}$$

$$\dfrac{GMm}{6R}$$

$$\dfrac{GMm}{24R}$$

Solution

Question 4

$$2 mgR$$

$$mgR$$

$$\dfrac{mgR}{4}$$

$$\dfrac{mgR}{2}$$

Solution

Question 5

$$\dfrac{GMm}{12R^{2}}$$

$$\dfrac{GMm}{3R^{2}}$$

$$\dfrac{GMm}{8R}$$

$$\dfrac{GMm}{6R}$$

Solution

Question 6

$$\dfrac{mgR}{2}$$

$$\dfrac{2mgR}{3}$$

$$\dfrac{3mgR}{4}$$

$$\dfrac{mgR}{3}$$

Solution

Question 7

$$2mgR$$

$$\dfrac{1}{2}mgR$$

$$\dfrac{1}{4}mgR$$

$$mgR$$

Solution

Question 8

$$\dfrac{W}{5}$$

$$\dfrac{W}{3}$$

$$\dfrac{W}{2}$$

$$\dfrac{W}{6}$$

Solution

Question 9

$$m$$

$$m^{-1}$$

$$m^{0}$$

$$m^{\frac{1}{2}}$$

Solution

Question 10

$$-3 J / kg$$

$$-2 J / kg$$

$$-5 J / kg$$

$$-7 J / kg$$

Solution

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