Gravitation Test

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Gravitation Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

Read the assertion and reason carefully to mark the correct option out of the options given below :

Assertion : At height $$h$$ from ground and at depth $$h$$ below ground, where h is approximately equal to $$0.62 R$$, the value of $$g$$ acceleration due to gravity is same.

Reason : Value of $$g$$ decreases both sides, in going up and down.

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

Inside the earth's surface, gravitational field varies as

$$g=\dfrac { GM }{ { r }^{ 2 } } \dfrac { { r }^{ 3 } }{ { R }^{ 3 } } =\dfrac { GMr }{ { R }^{ 3 } } $$

Outside the earth's surface,

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

For gravity at height and depth h to be same,

$$\dfrac { GM(R-h) }{ { R }^{ 3 } } =\dfrac { GM }{ { (R+h) }^{ 2 } } $$

One solution is $$h=0. $$Solving this, we get $$h\approx 0.62R$$

It is clear from both the expressions that the gravitational field decreases as one moves away from earth in any direction.

Answer is option A.
Question 2
1 / -0

A planet revolves around the sun in an elliptical orbit of eccentricity e. If T is the time period of the planet, then the time spent by the planet between the ends of the minor axis and major axis close to the sun is

A
$$\dfrac{T\pi}{2e}$$

B
$$T\left(\dfrac{2e}{\pi}-1\right)$$

C
$$\dfrac{Te}{2\pi}$$

D
$$T\left(\dfrac{1}{4}-\dfrac{e}{2\pi}\right)$$

Solution

As areal velocity of a planet around the sun is constant. Therefore, the desired time is

$${t}_{AB}=\left(\dfrac{area ABS}{area \ of \ ellipse}\right)\times time \ period$$

If a = semi-major axis and b= semi-minor axis of ellipse then, area of ellipse = $$\pi$$ab
Area $$ABS=\dfrac{1}{4}\left(area \ of \ ellipse \right)-Area \ of \ triangle ASO$$

$$=\dfrac{1}{4}\times\pi{ab}-\dfrac{1}{2}\left(ea\right)\times\left(b\right)$$

$$\therefore{t}_{AB}=\dfrac{\left[\dfrac{\pi\left(ab\right)}{4}-\dfrac{1}{2}eab\right]}{\pi ab}\times T=T\left(\dfrac{1}{4}-\dfrac{e}{2\pi}\right)$$
Question 3
1 / -0

Find out the correct relation for the dependance of change in acceleration due to gravity on the angle at the latitude due to rotation of earth?

A
$$dg\propto cos\phi $$

B
$$dg\propto { cos }^{ 2 }\phi $$

C
$$dg\propto { cos }^{ 3/2 }\phi $$

D
$$dg\propto \dfrac { 1 }{ cos\phi } $$

Solution

We know that, the effective gravity $$g'$$ at a latitude $$\phi$$ is given as
$$ g'=g-{ \omega }^{ 2 }R{ (\cos { \phi } ) }^{ 2 } $$
Thus, $$ g - g' = dg={ \omega }^{ 2 }R{ (\cos { \phi } ) }^{ 2 }$$
Question 4
1 / -0

The gravitational potential difference between the surface of a planet and a point $$20 m$$ above it is $$16 J/kg$$. Then the work done in moving a $$2 kg$$ mass by $$8 m$$ on a slope $$60$$ degree from the horizontal, is:

A
$$11.1 J$$

B
$$5.55 J$$

C
$$16 J$$

D
$$27.7 J$$
Question 5
1 / -0

At what height in km over the earth's pole the free fall acceleration decreases by one percent? (Assume the radius of the earth to be $$6400\ km$$)

A
$$32$$

B
$$64$$

C
$$80$$

D
$$1.253$$

Solution

Acceleration on the earth's surface is $$g=\dfrac{GM}{R^2}$$ and the acceleration at height h will be $$g_h=\dfrac{GM}{(R+h)^2}$$
So, $$\dfrac{g_h}{g}=\dfrac{R^2}{(R+h)^2}$$
According to question, $$\dfrac{g_h}{g}=\dfrac{99}{100}$$
so, $$\dfrac{99}{100}=\dfrac{R^2}{(R+h)^2}$$
or $$\dfrac{R}{(R+h)}=0.995$$ or $$R=0.995R+0.995h$$
or $$h=\left(\dfrac{0.005}{0.995}\right )R=\left(\dfrac{0.005}{0.995}\right)(6400)=32.16\simeq 32$$ km
Question 6
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Imagine a spacecraft going from the earth to the moon. How does its weight vary as it goes from the earth to the moon?

A
Initially decreases then becomes zero and again increases.

B
Initially increases and again increases.

C
Initially decreases again increases.

D
zero throughout the path

Solution

When a spacecraft is going from Earth to Moon,as long it is under Earth's gravitation,the acceleration due to gravity like $${ g }^{ \prime }={ g }(1-\cfrac { h }{ r } )$$ [g=acceleration due to gravity at the surface of the Earth]
Then, when it goes out of the Earth's gravitation,its weight will be zero.
After that,it will enter to the moon's gravitational field.
Then its acceleration due to Gravity will be $${ g }^{ \prime }={ g }(1-\cfrac { (-h) }{ r } )\quad $$
i.e,weight will be increasing
Question 7
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A body hanging from a massless spring stretches it by $$3$$cm on Earth's surface. At a place $$800$$km above the Earth's surface, the same body will stretch the spring by : (Radius of Earth $$=6400$$km)

A
$$\left(\displaystyle\frac{34}{27}\right)$$cm

B
$$\left(\displaystyle\frac{64}{27}\right)$$cm

C
$$\left(\displaystyle\frac{27}{64}\right)$$cm

D
$$\left(\displaystyle\frac{27}{34}\right)$$cm

E
$$\left(\displaystyle\frac{35}{81}\right)$$cm

Solution

Acceleration due to gravity,
$$g=\displaystyle\frac{GM}{r^2}$$
$$\therefore g\propto \frac{1}{r^2}$$
When the body is hanged on a spring
$$F=-kx=mg$$ ............(i)
Spring force $$F$$ acts on it, where x is extension in spring. Let $$800$$km above the Earth's surface, the stretch in the length of spring is $$x$$' and value of $$g$$ is $$g'$$.
$$g=\displaystyle\frac{GM}{R^2}$$ and $$g'=\displaystyle\frac{GM}{(R+h)^2}$$
So, $$g'=\displaystyle\frac{g}{\left(\displaystyle 1+\frac{h}{R}\right)^2}$$ .........(ii)
where, R=radius of Earth
From Eq. (i), we get
$$kx=mg$$
$$\Rightarrow x\propto g$$ .........(iii)
Therefore, $$\displaystyle\frac{x'}{x}=\frac{g'}{g}$$
From Eqs. (ii) and (iii), we get
$$\displaystyle\frac{x'}{x}=\frac{g}{g\left(\displaystyle 1+\frac{h}{R}\right)^2}=\frac{g}{\displaystyle g\left(1+\displaystyle\frac{800\times 10^3}{6400\times 10^3}\right)^2}$$
$$=\displaystyle \frac{g}{g\left(1+\displaystyle\frac{1}{8}\right)^2}$$
$$=\displaystyle\frac{g}{g\left(\displaystyle \frac{g}{8}\right)^2}=\frac{g\times 64}{g\times 81}=\frac{64}{81}$$
$$x'=\displaystyle\frac{64}{81}x$$ $$[x=3cm]$$
$$\Rightarrow x'=\displaystyle\frac{64}{81}\times 3=\frac{64}{27}$$
$$\Rightarrow x'=\displaystyle\frac{64}{27}$$cm.
Question 8
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A body is projected upwards with a velocity of $$4\times 11.2 km/s$$ from the surface of earth. What will be the velocity of the body when it escapes from the gravitational pull of earth?

A
$$11.2\ km/s$$

B
$$2\times 11.2\ km/s$$

C
$$3\times 11.2\ km/s$$

D
$$\sqrt {15}\times 11.2\ km/s$$

Solution
Question 9
1 / -0

A space vehicle approaching a planet has a speed v, when it is very far from the planet. At that moment tangent of its trajectory would miss the centre of the planet by distance R. If the planet has mass M and radius r, what is the smallest value of R in order that the resulting orbit of the space vehicle will just miss the surface of the planet?

A
$$\dfrac{r}{v}\left[v^2+\dfrac{2GM}{r}\right]^{\dfrac{1}{2}}$$

B
$$vr\left[1+\dfrac{2GM}{r}\right]$$

C
$$\dfrac{r}{v}\left[v^2+\dfrac{2GM}{r}\right]$$

D
$$\dfrac{2GMv}{r}$$

Solution
Question 10
1 / -0

A projectile is fired vertically upwards from the surface of earth with a velocity of $$kv_{e}$$ where $${v}_{e}$$ is the escape velocity and $$K<1$$. Neglecting air resistance, the maximum height to which it will rise, measured from the centre of the earth, is ($$R_{E}$$-radius earth)

A
$$\dfrac{R_{E}}{1-k^{2}}$$

B
$$\dfrac{R_{E}}{k^{2}}$$

C
$$\dfrac{1-k^{2}}{R_{E}}$$

D
$$\dfrac{k^{2}}{R_{E}}$$

Solution

Let a projectile is fired vertically upwards from the earth's surface with a velocity v and reaches a height h. Energy of the projectile at the surface of the earth is $$E_{i}=\dfrac{1}{2}mv^{2}-\dfrac{GM_{E}m}{R_{E}}$$
Energy of the projectile at a height h $$E_{f}=-\dfrac{GM_{E}m}{R_{E}+h}$$ ($$\therefore$$ At height h velocity of the projectile is zero.)
According to law of conservation of energy, $$E_{i}=E_{f}$$

$$\dfrac{1}{2}mv^{2}-\dfrac{GM_{E}m}{R_{E}}=-\dfrac{GM_{E}m}{R_{E}+h}$$

$$\dfrac{1}{2}mv^{2}=GM_{E}m\left[\dfrac{1}{R_{E}}-\dfrac{1}{R_{E}+h}\right]=\dfrac{GM_{E}mh}{\left(R_{E}\right)\left(R_{E}+h\right)}$$

$$\dfrac{1}{2}mv^{2}=\dfrac{mghR_{E}}{R_{E}+h} \left(\because g=\dfrac{GM_{E}}{R_{E}^{2}}\right)$$

As per question,

$$v=kv_{e}=k\sqrt{2gR_{E}} \left(\because v_{e}=\sqrt{2gR_{E}}\right)$$

and $$h=r-R_{E}$$

$$\therefore\dfrac{1}{2}mk^{2}2gR_{E}=\dfrac{mg\left(r-R_{E}\right)R_{E}}{r}$$

$$k^{2}=\dfrac{r-R_{E}}{r}\Rightarrow\left(1-k^{2}\right)r=R_{E}$$

$$\Rightarrow r=\dfrac{R_{E}}{1-k^{2}}$$

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Gravitation Test - 75

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Gravitation Test - 75

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SHARING IS CARING

Question 1

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 2

$$\dfrac{T\pi}{2e}$$

$$T\left(\dfrac{2e}{\pi}-1\right)$$

$$\dfrac{Te}{2\pi}$$

$$T\left(\dfrac{1}{4}-\dfrac{e}{2\pi}\right)$$

Solution

Question 3

$$dg\propto cos\phi $$

$$dg\propto { cos }^{ 2 }\phi $$

$$dg\propto { cos }^{ 3/2 }\phi $$

$$dg\propto \dfrac { 1 }{ cos\phi } $$

Solution

Question 4

$$11.1 J$$

$$5.55 J$$

$$16 J$$

$$27.7 J$$

Question 5

$$32$$

$$64$$

$$80$$

$$1.253$$

Solution

Question 6

Initially decreases then becomes zero and again increases.

Initially increases and again increases.

Initially decreases again increases.

zero throughout the path

Solution

Question 7

$$\left(\displaystyle\frac{34}{27}\right)$$cm

$$\left(\displaystyle\frac{64}{27}\right)$$cm

$$\left(\displaystyle\frac{27}{64}\right)$$cm

$$\left(\displaystyle\frac{27}{34}\right)$$cm

$$\left(\displaystyle\frac{35}{81}\right)$$cm

Solution

Question 8

$$11.2\ km/s$$

$$2\times 11.2\ km/s$$

$$3\times 11.2\ km/s$$

$$\sqrt {15}\times 11.2\ km/s$$

Solution

Question 9

$$\dfrac{r}{v}\left[v^2+\dfrac{2GM}{r}\right]^{\dfrac{1}{2}}$$

$$vr\left[1+\dfrac{2GM}{r}\right]$$

$$\dfrac{r}{v}\left[v^2+\dfrac{2GM}{r}\right]$$

$$\dfrac{2GMv}{r}$$

Solution

Question 10

$$\dfrac{R_{E}}{1-k^{2}}$$

$$\dfrac{R_{E}}{k^{2}}$$

$$\dfrac{1-k^{2}}{R_{E}}$$

$$\dfrac{k^{2}}{R_{E}}$$

Solution

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