Gravitation Test

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Gravitation Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

If R is the radius of the earth, then the ratio of acceleration due to gravity on surface of the earth to acceleration due to gravity at height nR is:

A
$$
\left( {n + 1} \right)^{ - 1}
$$

B
$$
\left( {n + 1} \right)^2
$$

C
$$
\left( {n + 1} \right)^{ - 2}
$$

D
$$
\left( {n + 1} \right)^3
$$

Solution

Given
Radius of Earth$$=R$$
Acceleration due to gravity $$g=\cfrac{GM}{R}$$
Also
$$g^{\prime}=g(\cfrac{1-2h}{R})$$
Given height $$=nR$$
$$g^{\prime}=g(\cfrac{1-2nR}{R})$$
When body is raised by $$nR$$
$$g^{\prime}=\cfrac{GM}{(R+nR)^2}\\g^{\prime}=\cfrac{GM}{R(1+n)^2}$$
Now ratio of $$g$$ and $$g^{\prime}$$
$$\Rightarrow(n+1)^2$$
Question 2
1 / -0

The value of $$g$$ on the earths surface is $$980\ cm/{sec}^{2}$$. Its value at a height of $$64\ Km$$ from the earth's surface is

A
$$960.40\ cm/sec ^{ 2 }$$

B
$$984.90\ cm/sec ^{ 2 }$$

C
$$982.45\ cm/sec ^{ 2 }$$

D
$$977.55\ cm/sec ^{ 2 }$$

Solution

At earth's surface. $$g=\dfrac{GM}{R^2}=9.8 m/s^2$$

$$\dfrac{GM}{(6400 \times 10^3)^2}=9.8 m/s^2$$ ...(1)

At a height 64 kms from earth's surface,

$$\dfrac{GM}{((6400+64) \times 10^3)^2}=g'$$ ...(2)

(1) $$\div$$ (2)

$$(\dfrac{6464}{6400})^2=\dfrac{9.8}{g'}$$

$$g'=(\dfrac{6464}{6400})^2 \times 9.8=9.6069 m/s^2$$

$$=960.69 cm/s^2$$
Question 3
1 / -0

The largest and the shortest distance of the earth from the sun are $$r_{1}$$ and $$r_{2}$$. Its distance from the sun when it is at perpendicular to the major-axis of the orbit drawn from the sun is:

A
$$\dfrac {r_{1}+r_{2}}{4}$$

B
$$\dfrac {r_{1}+r_{2}}{ r_{1}-r_{2}}$$

C
$$\dfrac {2r_{1}r_{2}}{ r_{1}+r_{2}}$$

D
$$\dfrac {r_{1}+r_{2}}{3}$$

Solution

From the property of ellipse
$$\dfrac{2}{R}=\dfrac{1}{r_1}+\dfrac{1}{r_2}$$
$$\dfrac{2}{R}=\dfrac{r_1+r_2}{r_1r_2}$$
$$R=\dfrac{2r_1r_2}{r_1+r_2}$$
Question 4
1 / -0

The height of the point vertically above the earths surface at which the acceleration due to gravity becomes $$1\%$$ of its value at the surface is ($$R$$ is the radius of the earth)

A
$$9R$$

B
$$10R$$

C
$$8R$$

D
$$20R$$

Solution
Question 5
1 / -0

At what height from the ground will be the value of $$g$$ be the same as
that in $$10 km$$ deep mine below the surface of earth.

A
20 km

B
7.5 km

C
5 km

D
2.5 km

Solution

Same changes in value of $$'g'$$ can be observe at a depth $$'x'$$
and height $$'2x'$$
given that the depth $$'d'=x=10km$$
$$\therefore$$ the same charge can be observed at a height $$2x=20km$$
Hence,
the correct option is $$A.$$
Question 6
1 / -0

A particle hanging from a spring stretches it by 1 cm at earth's surface. Radius of the earth is 6400 km. At a place 800 km above the earth's surface, the same particle will stretch the spring by

A
0.79 cm

B
1.2 cm

C
4 cm

D
17 cm

Solution

The extension is $$x=\dfrac{mg}{k}$$
At earth's surface, $$x=\dfrac{1}{100}=\dfrac{m \times \dfrac{GM}{R^2} }{k}$$..(1)
At 800 km,above the earth's surface,
$$x'=\dfrac{n\times \dfrac{GM}{(R+800)^2}}{k}$$ ..(2)
(2) $$\div$$ (1)
$$\dfrac{x'}{\dfrac{1}{100}}=\dfrac { \dfrac { GM }{ { \left( R+800 \right) }^{ 2 } } }{ \dfrac { GM }{ { R }^{ 2 } } } =\dfrac { { R }^{ 2 } }{ { \left( R+800 \right) }^{ 2 } }$$
$$\Rightarrow x'\times 100=\dfrac { x' }{ { \left( 1+\dfrac { 800 }{ R } \right) }^{ 2 } } =\dfrac { x' }{ { \left( 1+\dfrac { 800 }{ 6400 } \right) }^{ 2 } } =\dfrac { 1 }{ 1.265 } \\ =0.79\times { 10 }^{ -2 }\quad m\\ =0.79\quad cm$$
Question 7
1 / -0

The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 .(g being acceleration due to gravity at the surface of the earth). Maximum possible distance in terms of radius of earth R between P and Q is:

A
$$2R$$

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

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

D
$$\dfrac{R}{2} (2\sqrt{2}+1)$$

Solution

In above the surface,
$$\dfrac{g}{2} = \dfrac{GM}{(R+h)^2} ....1$$
And for below the surface,
$$\dfrac{g}{2} = \dfrac{GM}{(R-d)^2} ....2$$
Equating both the above equation then we get,
$$R^2 = h^2 + 2Rh$$
And,
$$R^2 = d^2 – 2Rd$$
By solving quadratic for both hand d terms as:
$$h = (\sqrt2 – 1)R$$
And,
$$d = (\sqrt2 + 1)R$$
Thus,
$$h + d = (\sqrt2 + 1)R + (\sqrt2 - 1)R$$
$$ => 2\sqrt{2}R$$ meter.
Question 8
1 / -0

A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be :

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

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

C
$$\dfrac{4W}{9}$$

D
$$\dfrac{W}{4}$$

Solution

Given, $$R=6400km,w=mg\Rightarrow g=\dfrac{w}{m}$$

Now the weight of the body at the height is $$\dfrac{1}{2}R$$

$$g'=g(1-\dfrac{d}{R})$$

So the weight of the body at the $$\dfrac{1}{2} R$$

$$W'=W(1-\dfrac{R}{2R})=W\dfrac{R}{2R})=\dfrac{W}{2}$$
Question 9
1 / -0

The height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (Radius of the earth = 6400 km) is (approximately) :

A
2630

B
2640

C
2650

D
2660

Solution

The new value of $$gravity$$ at a height $$h$$ is given as $$g\prime=\dfrac{gR^2}{(R+h)^2}$$
As visible from the formula that incresing $$h$$ results in decresing gravity.
So just put $$g\prime=g/2$$ i.e. 50% less than $$g$$ where $$g$$ is gravity at surface. so we get $$ g/2= \dfrac{gR^2}{(R+h)^2}$$ or $$ \dfrac{1}{2}= \dfrac{R^2}{(R+h)^2}$$
or $$\dfrac{2}{1}= \dfrac{(R+h)^2}{R^2}$$ or $$\sqrt[2]{2}= \dfrac{R+h}{R}=1+ h/R$$ putting values we get $$\dfrac{h}{R}= 1.414-1=0.414$$

so $$h=0.414R=2649.6Km$$
Option C is correct.
Question 10
1 / -0

The ratio of acceleration due to gravity at a depth $$h$$ below the surface of earth and at a height $$h$$ above the surface for $$h<< R$$

A
constants only when $$h<< R$$

B
increases linearly with $$h$$

C
increases parabolically with $$h$$

D
decreases

Solution

Above the surface, $$g'_{above}=\dfrac{GM}{(R+h)^2}$$ (at depth h)
Inside the earth, $$g'_{below}=\dfrac{GM}{R^2}(1-\dfrac{h}{R})$$
$$\dfrac{g'_{below}}{g'_{above}}=\dfrac{\dfrac{1}{R^2} (1-\dfrac{h}{R})}{\dfrac{1}{(R+h)^2}}=\dfrac { 1-\dfrac { h }{ R } }{ \dfrac { 2 }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } } $$
But for $$\dfrac { h }{ R } <<<1$$
$$\dfrac{g'_{below}}{g'_{above}}$$=$$(1-\dfrac { h }{ R } ){ \left( 1+\dfrac { h }{ R } \right) }^{ 2 }$$
$$=(1-\dfrac { h }{ R } ){ \left( 1+\dfrac { 2h }{ R } +{ \left( \dfrac { h }{ R } \right) }^{ 2 } \right) }\\ =1+\dfrac { 2h }{ R } +{ \left( \dfrac { h }{ R } \right) }^{ 2 }-\dfrac { h }{ R } -2{ \left( \dfrac { h }{ R } \right) }^{ 2 }-{ \left( \dfrac { h }{ R } \right) }^{ 3 }\\ =1+\dfrac { h }{ R } $$
It increases linearly with $$'h'$$

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 54

Result

Gravitation Test - 54

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SHARING IS CARING

Question 1

$$\left( {n + 1} \right)^{ - 1} $$

$$\left( {n + 1} \right)^2 $$

$$\left( {n + 1} \right)^{ - 2} $$

$$\left( {n + 1} \right)^3 $$

Solution

Question 2

$$960.40\ cm/sec ^{ 2 }$$

$$984.90\ cm/sec ^{ 2 }$$

$$982.45\ cm/sec ^{ 2 }$$

$$977.55\ cm/sec ^{ 2 }$$

Solution

Question 3

$$\dfrac {r_{1}+r_{2}}{4}$$

$$\dfrac {r_{1}+r_{2}}{ r_{1}-r_{2}}$$

$$\dfrac {2r_{1}r_{2}}{ r_{1}+r_{2}}$$

$$\dfrac {r_{1}+r_{2}}{3}$$

Solution

Question 4

$$9R$$

$$10R$$

$$8R$$

$$20R$$

Solution

Question 5

20 km

7.5 km

5 km

2.5 km

Solution

Question 6

0.79 cm

1.2 cm

4 cm

17 cm

Solution

Question 7

$$2R$$

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

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

$$\dfrac{R}{2} (2\sqrt{2}+1)$$

Solution

Question 8

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

$$\dfrac{2W}{3}$$

$$\dfrac{4W}{9}$$

$$\dfrac{W}{4}$$

Solution

Question 9

2630

2640

2650

2660

Solution

Question 10

constants only when $$h<< R$$

increases linearly with $$h$$

increases parabolically with $$h$$

decreases

Solution

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$$
\left( {n + 1} \right)^{ - 1}
$$

$$
\left( {n + 1} \right)^2
$$

$$
\left( {n + 1} \right)^{ - 2}
$$

$$
\left( {n + 1} \right)^3
$$