Gravitation Test

Result

Gravitation Test - 15

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

A planet in a distant solar system is $$10$$ times more massive than the earth and its radius is $$10$$ times smaller. Given that the escape velocity from the earth is $$11 km$$ $$\mathrm{s}^{-1}$$, the escape velocity from the surface ofthe planet would be

A
$$110 km \mathrm{s}^{-1}$$

B
$$0.11 km \mathrm{s}^{-1}$$

C
$$1.1 km \mathrm{s}^{-1}$$

D
$$11 km \mathrm{s}^{-1}$$

Solution

$$Escape\ velocity= \sqrt { \dfrac { 2Gm }{ r } } \\ \dfrac { { v }_{ new } }{ { v }_{ earth } } = \sqrt { \dfrac { { m }_{ new }{ r }_{ earth } }{ { m }_{ earth }{ r }_{ new } } } = \sqrt { \dfrac { 10 }{ \dfrac { 1 }{ 10 } } }= 10\\ { v }_{ new }= 110\ km{ s }^{ -1 }$$
Question 2
1 / -0

The mass of a spaceship is $$1000\ kg$$. It is to be launched from the earths surface out into free space. The value of $$g$$ and $$R$$ (radius of earth) are $$10\ m/s$$ and $$6400\ km$$ respectively. The required energy for this work will be :

A
$$6.4\times 10^{11}\ J$$

B
$$6.4\times 10^{8}\ J$$

C
$$6.4\times 10^{9}\ J$$

D
$$6.4\times 10^{10}\ J$$

Solution

$$\displaystyle W= \frac{GMm}{r}$$
$$\displaystyle =gR^2 \times \frac{m}{R}= mgR$$
$$=1000 \times 10 \times 6400 \times 10^3$$
$$=6.4 \times 10^{10}$$
Question 3
1 / -0

The height at which the acceleration due to gravity becomes $$\displaystyle \frac{\mathrm{g}}{9}$$ (where $$\mathrm{g}=$$ the acceleration due to gravity on the surface of the earth) in terms of $$\mathrm{R}$$, the radius of the earth, is:

A
$$2\ \mathrm{R}$$

B
$$\displaystyle \frac{\mathrm{R}}{\sqrt{2}}$$

C
$$\mathrm{R}/2$$

D
$$\sqrt{2}\ \mathrm{R}$$

Solution

$$\displaystyle \frac{\mathrm{G}\mathrm{M}}{9\mathrm{R}^{2}}=\frac{\mathrm{G}\mathrm{M}}{(\mathrm{R}+\mathrm{h})^{2}}$$
$$\Rightarrow 3\mathrm{R}=\mathrm{R}+\mathrm{h}$$
$$\Rightarrow \mathrm{h}=2\mathrm{R}$$
Question 4
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
$$2\mathrm{m}\mathrm{g}\mathrm{R}$$

B
$$\displaystyle \dfrac{1}{2}\mathrm{m}\mathrm{g}\mathrm{R}$$

C
$$\displaystyle \dfrac{1}{4}\mathrm{m}\mathrm{g}\mathrm{R}$$

D
$$\mathrm{m}\mathrm{g}\mathrm{R}$$

Solution

Acceleration due to gravity on earth's surface$$=g=\dfrac{GM}{R^2}$$.
Gain in potential energy=$$\dfrac{GMm}{R}-\dfrac{GMm}{2R}=\dfrac{GMm}{2R}=\dfrac{1}{2}mgR$$
Question 5
1 / -0

What is the minimum energy required to launch a satellite of mass $$m$$ from the surface of a planet of mass $$M$$ and radius $$R$$ in a circular orbit at an altitude of $$2R$$?

A
$$\displaystyle \frac{2GmM}{3R}$$

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

C
$$\displaystyle \frac{GmM}{3R}$$

D
$$\displaystyle \frac{5GmM}{6R}$$

Solution

Given that,
Mass of satellite $$=m$$
Mass of planet $$=M$$
Radius $$=R$$
Altitude $$h=2R$$
Now,
The gravitational potential energy
$$P.E=\dfrac{-Gm}{r}$$
Potential energy at altitude $$=\dfrac{GmM}{3R}$$
Orbital velocity $${{v}_{0}}=\sqrt{\dfrac{2GmM}{R+h}}$$

Now, the total energy is
$$ {{E}_{f}}=\dfrac{1}{2}mv_{0}^{2}-\dfrac{GmM}{3R} $$
$$ {{E}_{f}}=\dfrac{1}{2}\dfrac{GmM}{3R}-\dfrac{GMm}{3R} $$
$$ {{E}_{f}}=\dfrac{GmM}{3R}\left[ \dfrac{1}{2}-1 \right] $$
$$ {{E}_{f}}=\dfrac{-GmM}{6R} $$
Now, $${{E}_{i}}={{E}_{f}}$$
Now, the minimum required energy
$$ K.E=\dfrac{Gmm}{R}-\dfrac{GmM}{6R} $$
$$ K.E=\dfrac{5GmM}{6R} $$
Hence, the minimum required energy is $$\dfrac{5GmM}{6R}$$
Question 6
1 / -0

lf $$\mathrm{W}_{1},\ \mathrm{W}_{2}$$ and $$\mathrm{W}_{3}$$ represent the work done in moving a particle from A to $$\mathrm{B}$$ along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass $$\mathrm{m}$$, find the correct relation between $$\mathrm{W}_{1},\ \mathrm{W}_{2}$$ and $$\mathrm{W}_{3}$$.

A
$$\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}$$

B
$$\mathrm{W}_{1}=\mathrm{W}_{2}=\mathrm{W}_{3}$$

C
$$\mathrm{W}_{1}<\mathrm{W}_{2}<\mathrm{W}_{3}$$

D
$$\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}$$

Solution
Question 7
1 / -0

A body weighs $$72 N$$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

A
16 N.

B
28 N

C
32 N

D
72 N

Solution

Hint: Use formula of acceleration due to gravity at height h
Explanation:
Weight of the body on earth is given 72 N which is product of mass and acceleration due to gravity.
So, $$mg = 72$$.
Acceleration due to gravity at height h is given by the formula,
$$g’ = g \dfrac{g{R_E}^2}{{{(R}_E+\dfrac{R_E}{2})}^2}$$, where $$R_E$$ is radius of earth.
$$g’ = \dfrac{4}{9} g$$
So, weight on height h will be, $$W’ = mg’ = 72 * \dfrac{4}{9} = 32 N$$
$$Answer:$$
Hence, option C is the correct answer.
Question 8
1 / -0

A body of mass '$$m$$' taken from the earth's surface to the height equal to twice the radius ($$R$$) of the earth. The change in potential energy of body will be

A
$$2mgR$$

B
$$\displaystyle \frac{1}{3}mgR$$

C
$$3mgR$$

D
$$\displaystyle \frac{2}{3}mgR$$

Solution
Question 9
1 / -0

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

A
$$150 N $$

B
$$200 N$$

C
$$250 N$$

D
$$100 N $$

Solution

Acceleration due to gravity at a depth d from surface of earth
$$g' = g \left(1 - \dfrac{d}{R} \right) $$ ....(1)
Where g = acceleration due to gravity at earth's surface
Multiplying by mass 'm' on both sides of (1)
$$\implies mg' = mg \left(1 - \dfrac{d}{R} \right) \, \,\, \, Here, \left(d = \dfrac{R}{2} \right)$$
$$= 200 \left(1 - \dfrac{R}{2 R} \right) = \dfrac{200}{2} = 100 N$$
Question 10
1 / -0

The ratio of escape velocity at earth$$(v_e)$$ to the escape velocity at a planet$$(v_p)$$ whose radius and mean density are twice as that of earth is:

A
$$1:2$$

B
$$1:2\sqrt 2$$

C
$$1:4$$

D
$$1:\sqrt 2$$

Solution

We know that: $$v_e=\sqrt{\dfrac{2GM}{R}}$$

Given that: $$R'=2R$$ and $$\rho'=2\rho$$

$$\Rightarrow M=\dfrac{4}{3}\pi R^3\times \rho$$

$$\Rightarrow M'=\dfrac{4}{3}\pi(2R)^3\times 2\rho=16M$$

$$\Rightarrow v_p=\sqrt{\dfrac{2G(16M)}{2R}}=\sqrt{\dfrac{16GM}{R}}$$

$$\Rightarrow \dfrac{v_e}{v_p}=\sqrt{\dfrac{1}{8}}=\dfrac{1}{2\sqrt{2}}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Gravitation Test - 15

Result

Gravitation Test - 15

Score

-

Rank

-

SHARING IS CARING

Question 1

$$110 km \mathrm{s}^{-1}$$

$$0.11 km \mathrm{s}^{-1}$$

$$1.1 km \mathrm{s}^{-1}$$

$$11 km \mathrm{s}^{-1}$$

Solution

Question 2

$$6.4\times 10^{11}\ J$$

$$6.4\times 10^{8}\ J$$

$$6.4\times 10^{9}\ J$$

$$6.4\times 10^{10}\ J$$

Solution

Question 3

$$2\ \mathrm{R}$$

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

$$\mathrm{R}/2$$

$$\sqrt{2}\ \mathrm{R}$$

Solution

Question 4

$$2\mathrm{m}\mathrm{g}\mathrm{R}$$

$$\displaystyle \dfrac{1}{2}\mathrm{m}\mathrm{g}\mathrm{R}$$

$$\displaystyle \dfrac{1}{4}\mathrm{m}\mathrm{g}\mathrm{R}$$

$$\mathrm{m}\mathrm{g}\mathrm{R}$$

Solution

Question 5

$$\displaystyle \frac{2GmM}{3R}$$

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

$$\displaystyle \frac{GmM}{3R}$$

$$\displaystyle \frac{5GmM}{6R}$$

Solution

Question 6

$$\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}$$

$$\mathrm{W}_{1}=\mathrm{W}_{2}=\mathrm{W}_{3}$$

$$\mathrm{W}_{1}<\mathrm{W}_{2}<\mathrm{W}_{3}$$

$$\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}$$

Solution

Question 7

16 N.

28 N

32 N

72 N

Solution

Question 8

$$2mgR$$

$$\displaystyle \frac{1}{3}mgR$$

$$3mgR$$

$$\displaystyle \frac{2}{3}mgR$$

Solution

Question 9

$$150 N $$

$$200 N$$

$$250 N$$

$$100 N $$

Solution

Question 10

$$1:2$$

$$1:2\sqrt 2$$

$$1:4$$

$$1:\sqrt 2$$

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.