Gravitation Test

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Gravitation Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is: ($$g$$=acceleration due to gravity near the poles and $$R$$ is the radius of earth) (Ignore equatorial bulge)

A
$$\displaystyle 2\pi \sqrt { \frac { R }{ 2g } } $$

B
$$\displaystyle 2\pi \sqrt { \frac { R }{ 3g } } $$

C
$$\displaystyle 2\pi \sqrt { \frac { R }{ g } } $$

D
$$\displaystyle 2\pi \sqrt { \frac { 2R }{ g } } $$

Solution

Let weight of the person be $$mg$$ and the angular speed of the earth be $$\omega$$.
Thus weight of the person at poles=$$mg$$
Weight of the person at equator=$$mg-m\omega ^2 R$$
$$\implies mg=2(mg-m\omega ^2R)$$
$$\implies \omega ^2R=\dfrac{g}{2}$$
$$\implies \omega=\sqrt{\dfrac{g}{2R}}$$
Thus time period of earth's rotation=$$\dfrac{2\pi}{\omega}$$
$$=2\pi\sqrt{\dfrac{2R}{g}}$$
Question 2
1 / -0

When a body is at a depth 'd' from the earth surface its distance from the centre of the earth is _______

A
$$(R - d)$$

B
$$2(R - d)$$

C
$$(3R - d)$$

D
$$(R - 2d)$$

Solution

$$SP=d$$
$$SQ=r$$
$$PQ=(R-d)=$$ Distance of the from the centre of the Earth.
Question 3
1 / -0

If R is the radius of earth, the height at which the weight of a body becomes $$1/4$$ its weight on the surface of earth is :

A
$$2R$$

B
$$R$$

C
$$R/2$$

D
$$R/4$$

Solution

At surface of the Earth, weight $$W_s=\dfrac{GMm}{R^2}$$
At height, h of the Earth weight $$W_h=\dfrac{GMm}{(R+h)^2}$$
Now,
$$W_h = \dfrac{W_s}{4}$$

$$\dfrac{GMm}{(R+h)^2} = \dfrac{GMm}{4R^2}$$

$$\dfrac{1}{R+h}=\dfrac{1}{2R}$$

$$R+h=2R$$

$$h=R$$
Question 4
1 / -0

Every planet revolves around the sun in a/an ______ orbit.

A
elliptical

B
circular

C
parabolic

D
none of these

Solution

The first law of Kepler states that the orbit of a planet is an ellipse with the Sun at one of the two foci.

Hence correct answer is option $$A $$
Question 5
1 / -0

The ratio of SI unit of G to its CGS unit is _______.

A
$$100 : 1$$

B
$$1000 : 1$$

C
$$10 : 1$$

D
$$10000 : 1$$

Solution

We know that SI unit of G $$=m^3/s^2kg=(100cm) ^3/s^2(1000g)=10^6cm^3/10^3gs^2=10^3cm^3/gs^2$$

Also the CGS unit of G $$=cm^3/gs^2$$
$$\Rightarrow \dfrac{SI}{CGS}=1000$$

Hence correct answer is option $$B$$
Question 6
1 / -0

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

A
$$72 N$$

B
$$28 N$$

C
$$16 N$$

D
$$32 N$$

Solution

Hint:- Use the formula of gravitational acceleration to check weight of the body on certain height.

Step-1 Note the given values
Let $$R_e$$ be the radius of earth
Given
height, $$h = \frac{R_e}2$$
Acceleration due to gravity at the surface of earth=g

Step -2 Calculate gravitational force on height h
Since h is comparable to $$R_e$$
So, $$g_h=\dfrac{GM}{(R_e+h) ^2}$$
Putting $$h=R_e/2$$,we get
$$\Rightarrow g_h=\dfrac{4}{9}\times \dfrac{GM}{R_e^2}=\dfrac{4g}{9}$$

Therefore weight at a height equal to half of the earth's radius $$ mg_h=\dfrac{4}{9}\times mg= \dfrac{4}9 \times {72 N}=32 N$$

Hence correct answer is option $$D$$
Question 7
1 / -0

The figure given below shows a planet in elliptical orbit around the sun $$S$$. At what position will the kinetic energy of the planet be maximum?

A
$${ P }_{ 1 }$$

B
$${ P }_{ 4 }$$

C
$${ P }_{ 3 }$$

D
$${ P }_{ 2 }$$

Solution

Hint : Use angular momentum conservation

Step :1 Finding the point at which velocity is maximum.
As there is no external torque on the system so angular momentum is conserved.
L=mvr=constant; where m=mass of satellite,v=velocity of satellite, and r=distance of the satellite from Sun
From here we can see that $$v \propto \frac1r$$
V is maximum at the point where r is minimum
So velocity is maximum at point $$P_1$$

Step 2: Finding point of maximum kinetic energy
As kinetic energy ,$$K =\dfrac{1}{2}mv^2$$
Kinetic energy is maximum at the point where V is maximum
So kinetic energy is maximum at point $$P_1$$

$$\textbf{Hence option A correct}$$
Question 8
1 / -0

At what height from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be $$\dfrac {g}{2}?$$

A
$$R\sqrt {2}$$

B
$$R(\sqrt {2} - 1)$$

C
$$R$$

D
$$R(\sqrt {2} - 2)$$

Solution

We have, $${ g }_{ h }=\dfrac { g }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$

$${ g }_{ h }=\dfrac { g }{ 2 } $$

$$\therefore \dfrac { g }{ 2 } =\dfrac { g }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$

$$\Rightarrow \dfrac { 1 }{ 2 } =\dfrac { 1 }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$

$${ \left( 1+\dfrac { h }{ R } \right) }^{ 2 }=2$$

$$1+\dfrac { h }{ R } =\sqrt { 2 }$$

$$\dfrac { h }{ R } =\sqrt { 2 } -1$$

$$h=R\left( \sqrt { 2 } -1 \right) $$
Question 9
1 / -0

At what height from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be $$\dfrac {4g}{9}?$$

A
$$9R/4$$

B
$$R/2$$

C
$$3R/2$$

D
$$4R/9$$

Solution

We have, $${ g }_{ h }=\dfrac { g }{ { \left( 1+\cfrac { h }{ R } \right) }^{ 2 } } $$

$${ g }_{ h }=\dfrac { 4g }{ 9 } $$

$$\therefore \dfrac { 4g }{ 9 } =\dfrac { g }{ { \left( 1+\cfrac { h }{ R } \right) }^{ 2 } } \quad \Rightarrow 2\left( 1+\cfrac { h }{ R } \right) =3$$

$$\Rightarrow 1+\cfrac { h }{ R } =\cfrac { 3 }{ 2 } $$

$$\cfrac { h }{ R } =\cfrac { 3 }{ 2 } -1=\cfrac { 1 }{ 2 } $$

or, $$h=\cfrac { R }{ 2 } $$
Question 10
1 / -0

A body weighs $$100\ N$$ at a distance $$\dfrac {R}{4}$$ from centre of earth. Find its weight at height of $$9\ R$$ from the surface of earth (R - Radius of earth)

A
$$400\ N$$

B
$$3.6\ N$$

C
$$1\ N$$

D
$$4\ N$$

Solution

distance $$=\dfrac { R }{ 4 } $$
$$\therefore $$ Depth $$=d=\dfrac { 3R }{ 4 } $$
$${ g }_{ a }=g\left( 1-\dfrac { d }{ R } \right) $$
$$=g\left( 1-\dfrac { 3 }{ 4 } \right)$$
$$=\dfrac { g }{ 4 } $$
$$\therefore \quad 100N=\dfrac { mg }{ 4 } $$
or $$mg=400N$$
Also, $${ g }_{ n }=\dfrac { g }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$
Distance $$=9R$$
$$\therefore $$ Height, $$h=8R$$
$${ g }_{ h }=\dfrac { g }{ { \left( 1+\dfrac { 8R }{ R } \right) }^{ 2 } } \Rightarrow { g }_{ h }=\dfrac { g }{ { 9 }^{ 2 } } $$
$${ g }_{ h }=\dfrac { g }{ 81 } $$

or $$mg=81m{ g }_{ h }$$
$$\Rightarrow 400=81{ g }_{ h }m$$
or $$m{ g }_{ h }=\dfrac { 400 }{ 81 } \approx 4N$$

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Gravitation Test - 42

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Gravitation Test - 42

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SHARING IS CARING

Question 1

$$\displaystyle 2\pi \sqrt { \frac { R }{ 2g } } $$

$$\displaystyle 2\pi \sqrt { \frac { R }{ 3g } } $$

$$\displaystyle 2\pi \sqrt { \frac { R }{ g } } $$

$$\displaystyle 2\pi \sqrt { \frac { 2R }{ g } } $$

Solution

Question 2

$$(R - d)$$

$$2(R - d)$$

$$(3R - d)$$

$$(R - 2d)$$

Solution

Question 3

$$2R$$

$$R$$

$$R/2$$

$$R/4$$

Solution

Question 4

elliptical

circular

parabolic

none of these

Solution

Question 5

$$100 : 1$$

$$1000 : 1$$

$$10 : 1$$

$$10000 : 1$$

Solution

Question 6

$$72 N$$

$$28 N$$

$$16 N$$

$$32 N$$

Solution

Question 7

$${ P }_{ 1 }$$

$${ P }_{ 4 }$$

$${ P }_{ 3 }$$

$${ P }_{ 2 }$$

Solution

Question 8

$$R\sqrt {2}$$

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

$$R$$

$$R(\sqrt {2} - 2)$$

Solution

Question 9

$$9R/4$$

$$R/2$$

$$3R/2$$

$$4R/9$$

Solution

Question 10

$$400\ N$$

$$3.6\ N$$

$$1\ N$$

$$4\ N$$

Solution

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