Gravitation Test

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Gravitation Test - 37

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Question 1
1 / -0

The ratio of the radii of the planets $${P}_{1}$$ and $${P}_{2}$$ is $$k$$. The ratio of gravitational field intensity at their surface is $$r$$ then the ratio of the escape velocities from them will be -

A
$$kr$$

B
$$\sqrt{kr}$$

C
$$\displaystyle\sqrt{\frac{k}{r}}$$

D
$$\displaystyle\sqrt{\frac{r}{k}}$$

Solution

We are given that:
$$\dfrac{\dfrac { G{ M }_{ 1 } }{ { R }_{ 1 }^{ 2 } }} {\dfrac { G{ M }_{ 2 } }{ { R }_{ 2 }^{ 2 } } }=r$$

$$\\ \dfrac { { R }_{ 1 } }{ { R }_{ 2 } } =k$$.
And the escape velocity is given by: $$v=\sqrt { \dfrac { 2GM }{ R } }$$ where $$M$$ is the mass of the planet and $$R$$ is its radius.
So, $$\dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\sqrt { \dfrac { { M }_{ 1 } }{ { M }_{ 2 } } /\dfrac { { R }_{ 1 } }{ { R }_{ 2 } } } =\sqrt { \dfrac { r{ k }^{ 2 } }{ k } } =\sqrt { rk } $$.
Question 2
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Mass of a planet is $$5 \times 10^{24}$$ kg and radius is $$6.1 \times 10^6$$m. The energy needed to send a $$2$$ kg body into space from its surface, would be.

A
$$9$$ joule

B
$$18$$ joule

C
$$2.2 \times 10^8 $$joule

D
$$1.1 \times 10^8 $$joule

Solution

Energy required to send a body into the space from the surface of planet is equal to the negative of potential energy at surface.
At surface $$P.E = -\dfrac{GMm}{R}$$
Thus energy required $$E= \dfrac{GMm}{R}$$
Given: $$M= 5 \times 10^{24} kg m= 2 kg R= 6.1 \times 10^6 m G= 6.67 \times 10^{-11} Nm^2/kg^2$$
$$\implies E=1.1 \times 10^{8} Joule$$
Question 3
1 / -0

Gravitational potential difference between surface of a planet and a point situated at a height of $$20m$$ above its surface is $$2 joule/kg$$. If gravitational field is uniform, then the work done in taking a $$5kg$$ body of height $$4$$ meter above surface will be :-

A
$$2J$$

B
$$20J$$

C
$$40J$$

D
$$10J$$

Solution

The potential difference for a distance of $$20 m$$ is $$2 \ joule/kg$$.
Now work done $$=m \times$$ potential difference.
Thus work done for a distance of $$4 m$$ for a mass $$5 kg$$, $$W= \dfrac{4}{20} \times 2 \times 5$$
$$\implies W= 2 J$$
Question 4
1 / -0

An object weighs $$10 N$$ at the north pole of theearth. In a geostationary satellite distance $$7R$$ from the centre of the earth (of radius $$R$$), the true weight and the apparent weight are.

A
$$0 N, 0 N$$

B
$$0.2 N, 0 N$$

C
$$0.2 N, 9.8 N$$

D
$$0.2 N, 0.2 N$$

Solution

Acceleration due to gravity at a height h from the surface , $$g'= g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$$
For radius of orbit $$= 7R \implies h=6R$$
$$mg'= mg\bigg[1+\dfrac{6R}{R}\bigg]^{-2}$$
$$mg'= 10 \times [7]^{-2} N $$
$$\implies W'= .2 N$$
Also apparent weight of an object will bo zero.
Question 5
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A body of mass $$m$$ is situated at distance $$4R_e$$ above the earth's surface, where $$R_e$$ is the radius of earth how much minimum energy be given to the body so that it may escape

A
$$mgR_e$$

B
$$2mgR_e$$

C
$$\dfrac {mgR_e} {5}$$

D
$$\dfrac {mgR_e} {16}$$

Solution

Potential energy of the body at a distance 4$$R_e$$ from the surface of earth
$$U=-\dfrac{mgR_e}{1+n/R_e}$$

$$=-\dfrac{mgR_e}{1+4}=-\dfrac{mgR_e}{5} $$ As $$h=4R_e$$
So, minimum energy required to escape the body will be $$\dfrac{mgR_e}{5}$$
Question 6
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Read the assertion and reason carefully to mark the correct option out of the options given below :

Assertion : Radius of circular orbit of a satellite is made two times, then it areal velocity will also become two times.
Reason : Areal velocity is given as $$\dfrac {dA}{dt}=\dfrac {L}{2m}=\dfrac {mvr}{2m}$$

A
If both assertion and reason are true and the reason is the correct explanation of the assertion

B
If both assertion and reason are true but reason is not the correct explanation of the assertion

C
If assertion is true but reason is false

D
If assertion is false but reason is true

Solution

Areal velocity $$\dfrac{dA}{dt}= \dfrac{L}{2m}=\dfrac{mvr}{2m}$$
As there is no external torque, thus $$L=constant$$
Now if radius of orbit increases by 2 times then the velocity of satellite will get reduced by 2 times so as to make $$L$$ to be constant.
Hence overall, increasing $$r$$ has no effect on areal velocity.
So the assertion is false.
Question 7
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If the change in the value of '$$g$$' at a height $$h$$ above the surface of the earth is the same as at a depth $$d$$ below it. When both $$d$$ and $$h$$ are much smaller than the radius of earth, then

A
$$d = h$$

B
$$d = 2h$$

C
$$d = h/2$$

D
$$d = h$$$$^2$$

Solution

the value of 'g' at a height $$h$$ above the surface of the earth, $$g_h=g(1-\dfrac{2h}{R})$$
the value of 'g' at a depth d, $$g_d=g(1-\dfrac{d}{R})$$
But, $$g_h=g_d$$
$$\therefore$$ $$1-\dfrac{2h}{R}=1-\cfrac{d}{R}$$
or, $$d=2h$$
Question 8
1 / -0

The weight of an object would be minimum when it is placed :

A
At the North Pole

B
At the South Pole

C
At the Equator

D
At the Centre of the Earth

Solution

$$Answer:-$$ D
Value of g at any point on surface of earth is given by:-
$${ g }^{ \prime }=g-{ \omega }^{ 2 }R\cos { \varphi }$$
At equator $$cos\varphi =1\ so\ { g }^{ \prime }=g-{ \omega }^{ 2 }R $$
Value of $$\cos { \varphi } <1$$ at every point other than equator.
So, $$ { g }^{ \prime }$$ will be more at any point other than equator relative to equator.
Variation of $$g$$ with depth is given by:
$$g_d=g(1-\dfrac{d}{R})$$
So, when depth d is maximum g is minimum. So, weight (mg) will also be minimum.
Question 9
1 / -0

If $$A$$ is the areal velocity of planet of mass $$M$$. its angular momentum is

A
$$M$$

B
$$2MA$$

C
$$A^2M$$

D
$$AM^2$$

Solution

$$\textbf{Hint: First calculate $\omega$ and Moment of inertia of planet is, $I=M{R}^{2}$.}$$

$$\textbf{Step 1: Calculating the angular velocity.}$$

Given that the planet has an aerial velocity of A, Aerial velocity is given by,

$$A=\dfrac{Area swept by radius vector}{Time taken}$$

$$\Rightarrow A=\dfrac{\pi {{R}^{2}}}{T}$$ (Where R is the radius of the planet)

We know that, $$T=\dfrac{2\pi }{\omega }$$

$$\Rightarrow A=\dfrac{\omega {{R}^{2}}}{2}$$

$$\Rightarrow \omega =\dfrac{2A}{{{R}^{2}}}$$

$$\textbf{Step 2: Calculating the angular momentum.}$$

Angular momentum of the planet will be, $$L=I\omega =M{{R}^{2}}\times \dfrac{2A}{{{R}^{2}}}$$

$$\Rightarrow L=2MA$$
$$\textbf{Thus, option (B) is correct.}$$
Question 10
1 / -0

The velocity of a planet revolving around the sun at three different times of a year is shown in the figure. Which among the following alternatives is correct ?

A
$$v_2\, =\, \displaystyle \frac{v_1\, +\, v_3}{2}$$

B
$$v_1\, =\, \displaystyle \frac{v_2\, +\, v_3}{2}$$

C
$$v_1\, >\, v_2\, >\, v_3$$

D
$$v_1\, <\, v_2\, <\, v_3$$

Solution

Kepler's law states that angular momentum is conserved.
$$L = constant$$
$$v_{1} r_{1} = v_{2}r_{2} = v_{3}r_{3}$$
r is the position vector from one point to the sun.
As we can see from the ellipse:
$$r_{3} > r_{2} > r_{1}$$
Hence,
$$v_{1} > v_{2} > v_{3}$$
Option C is correct.

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Gravitation Test - 37

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Gravitation Test - 37

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SHARING IS CARING

Question 1

$$kr$$

$$\sqrt{kr}$$

$$\displaystyle\sqrt{\frac{k}{r}}$$

$$\displaystyle\sqrt{\frac{r}{k}}$$

Solution

Question 2

$$9$$ joule

$$18$$ joule

$$2.2 \times 10^8 $$joule

$$1.1 \times 10^8 $$joule

Solution

Question 3

$$2J$$

$$20J$$

$$40J$$

$$10J$$

Solution

Question 4

$$0 N, 0 N$$

$$0.2 N, 0 N$$

$$0.2 N, 9.8 N$$

$$0.2 N, 0.2 N$$

Solution

Question 5

$$mgR_e$$

$$2mgR_e$$

$$\dfrac {mgR_e} {5}$$

$$\dfrac {mgR_e} {16}$$

Solution

Question 6

If both assertion and reason are true and the reason is the correct explanation of the assertion

If both assertion and reason are true but reason is not the correct explanation of the assertion

If assertion is true but reason is false

If assertion is false but reason is true

Solution

Question 7

$$d = h$$

$$d = 2h$$

$$d = h/2$$

$$d = h$$$$^2$$

Solution

Question 8

At the North Pole

At the South Pole

At the Equator

At the Centre of the Earth

Solution

Question 9

$$M$$

$$2MA$$

$$A^2M$$

$$AM^2$$

Solution

Question 10

$$v_2\, =\, \displaystyle \frac{v_1\, +\, v_3}{2}$$

$$v_1\, =\, \displaystyle \frac{v_2\, +\, v_3}{2}$$

$$v_1\, >\, v_2\, >\, v_3$$

$$v_1\, <\, v_2\, <\, v_3$$

Solution

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