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CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Physics
Gravitation
Quiz 7
Weight of a body of mass m decreases by $$1 \%$$ when it is raised to height $$h$$ above the earth's surface. If the body is taken to a depth $$h$$ in a mine, change in its weight is:
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by $$0.5 \%$$ decrease
0%
by $$2 \%$$ decrease
0%
by $$0.5 \%$$ increase
0%
by $$1 \%$$ increase
Explanation
Hint:
The weight of a body is defined as the gravitational force exerted on the body.
Step 1: Calculate the density of the earth and write the formula of the gravitational force.
$$density=\dfrac{mass}{volume}$$
$$\rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}$$
Here,
$$M$$
is the mass,
and
$$R$$
is the radius of the earth.
The force between two masses can be expressed as,
$$F=\dfrac{GM_1M_2}{r^2}$$
Step 2: Find the value of $$h$$.
Let the mass of the earth be $$M$$ and the mass of the body is $$m$$.
The weight of the body on the earth is,
$$F=\dfrac{GMm}{R^2}$$
The weight of the body when it is at height $$h$$ is,
$$F_1=\dfrac{GMm}{(R+h)^2}$$
The weight is decreased by 1%, Hence,
$$\dfrac{F-F_1}{F} \times 100 =1$$
$$[1-\dfrac{\dfrac{GMm}{(R+h)^2}}{\dfrac{GMm}{R^2}}] \times 100 =1$$
$$\dfrac{(R+h)^2-R^2}{(R+h)^2}=\dfrac{1}{100}$$
$$\dfrac{R^2+h^2+2Rh-R^2}{(R+h)^2}=\dfrac{1}{100}$$
$$\because h<<R$$
$$\therefore {(R+h)}^2 \rightarrow R^2$$ and $$h^2$$ is very small, therefore it can be ignored.
$$\dfrac{2hR}{(R)^2}=\dfrac{1}{100}$$
$$h=\dfrac{R}{200}$$
Step 3: Finding the weight inside the earth at depth $$h$$
The weight of the body when it is at depth $$h$$ is,
$$F_2=\dfrac{GM_hm}{(R-h)^2}$$
Here $$M_h$$ is the effective mass of the earth at depth $$h$$.
$$mass=density\times volume$$
$$M_h=\rho \times \frac{4}{3}\pi (R-h)^3$$
$$M_h=\dfrac{M}{\dfrac{4}{3}\pi R^3} \times \frac{4}{3}\pi (R-h)^3$$
$$M_h=\dfrac{M}{R^3} \times(R-h)^3$$
Substituting this value in the expression of $$F_2$$
$$F_=\dfrac{GMm(R-h)}{(R)^3}$$
Step 4: Calculate the weight loss,
The weight loss can be calculated as,
$$\Delta W$$
$$=\dfrac{F-F_1}{F} \times 100 $$
$$\Delta W$$
$$=[1-\dfrac{\dfrac{GMm(R-h)}{(R)^3}}{\dfrac{GMm}{R^2}}] \times 100 $$
$$\Delta W$$
$$=[1-{\dfrac{(R-h)}{(R)}}] \times 100 $$
Substituting the value of $$h$$,
$$\Delta W =0.5%$$%
Therefore, the weight decreased by $$0.5$$%.
$$\textbf{Hence option A correct}$$
Calculate the value of $$g$$
on the surface of planet if t
he planet has $$1/500$$ the mass and $$1/15$$ the radius of the Earth.
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$$0.3 m/s_2$$
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$$1.6 m/s_2$$
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$$2.4 m/s_2$$
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$$4.5 m/s_2$$
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$$7.1 m/s_2$$
Explanation
The value of $$g$$ on the surface of planet$$=\dfrac{GM_{planet}}{R^2}$$
Value of $$g$$ on earth's surface$$=g=10m/s^2=\dfrac{GM_{earth}}{R_{earth}^2}$$
The value of $$g$$ on the planet$$=\dfrac{GM_{planet}}{R_{planet}^2}$$
$$=\dfrac{15^2}{500}\times g_{earth}=4.5m/s^2$$
An object is released from rest at a distance of $$2vr_e$$ from the center of the Earth, where $$r_e$$ is the radius of the Earth. In terms of the gravitational constant (G ), the mass of the Earth (M ), and $$r_e$$, what is the velocity of the object when it hits the Earth?
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$$\sqrt{GM/r_e}$$
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$$GM/r_e$$
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$$\sqrt{GM/2r_e}$$
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$$GM/2r_e$$
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$$2GM/r_e$$
Explanation
When the object falls from height $$2r_e$$ to $$r_e$$, the loss in gravitational potential energy would be equal to the gain in kinetic energy.
Hence $$-\dfrac{GMm}{2r_e}-(-\dfrac{GMm}{r_e})=\dfrac{1}{2}mv^2$$
$$\implies \dfrac{GMm}{2r_e}=\dfrac{1}{2}mv^2$$
$$\implies v=\sqrt{\dfrac{GM}{r_e}}$$
A person normally weighs 800N at sea level, climbs to the top of a mountain. While on top of the mountain that person will weigh:
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Zero
0%
Approximately $$800N$$
0%
Considerably less than $$800N$$ but more than zero
0%
Considerably more than $$800N$$
0%
Need more information
Explanation
When we go up from sea level at a height $$h$$ gravitational acceleration is given by the relation $$g'=g\left(1-2h/R \right)$$ where R is the radius of earth ,
as in this case h is negligible compared to R therefore g will not change considerably
and due to that weight $$mg$$ will decrease but negligibly by small amount. Therefore it will remain approximately $$800N$$.
A diver stands a top a platform that is $$15$$ meters high. After diving, She challenges herself from a cliff that is $$30$$ meters high. Since she is twice as far from the surface of the earth when she is on the cliff as compared with the diving board. how does her weight on the cliff compare with her weight on the diving board?
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Her weight on the cliff is half as much.
0%
Her weight on the cliff is one-fourth as much.
0%
Her weight on the cliff is about the same.
0%
Her weight on the cliff is twice as much.
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Her weight on the cliff is four times as much.
Explanation
Newton's law of universal gravitation helps us answer this question
$$F=G\dfrac { { { m }_{ 1 }m }_{ 2 } }{ { r }^{ 2 } }$$
The r in the denominator is the distance between centers of masses of objects -not between their surfaces
This is especially important to take into account when applying the equation to large objects like earth . the difference in distance between the diver and the center of the earth is insignificantly different when she is standing on the cliff compared to when she is standing on the diving board .
hence her weight is only very slightly different - about the same
so the correct choice is option "c"
Fifteen joules of work is done on object A so that only its gravitational potential energy changes. Sixty joules of work is done on object B (same mass as object A) so that only its gravitational potential energy changes.
How many times does the height of object B change compared to the height change of object A, as result of the work done?
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Object B changes height four times as much as object A changes height
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Object B changes height sixteen times as much as object A changes height
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Object B changes height two times as much as object A changes height
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Object B changes height less than two times as much as object A changes height (but not the same amount)
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Object B changes height the same amount as object A changes height
Explanation
Since the work done on object is stored as potential energy of object,
$$W=mg\Delta h$$
$$\implies \dfrac{W_A}{W_B}=\dfrac{\Delta h_A}{\Delta h_B}=\dfrac{15}{60}=\dfrac{1}{4}$$
$$\implies \Delta h_B=4\Delta h_A$$
Calculate the mass and weight of the object which is at height of twice the radius of the Earth from the surface of the Earth, if mass and weight of the object is $$m$$ and $$w$$ respectively.
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its mass is $$\dfrac{m}{2}$$ and its weight is $$\dfrac{w}{2}$$
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its mass is $$m$$ and its weight is $$\dfrac{w}{8}$$
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its mass is $$\dfrac{m}{2}$$ and its weight is $$\dfrac{w}{4}$$
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its mass is $$m$$ and its weight is $$\dfrac{w}{4}$$
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its mass is $$m$$ and its weight is $$\dfrac{w}{9}$$
Explanation
Weight of the object at the surface$$=w=\dfrac{GMm}{R^2}$$
Mass of the object is universally constant. Thus it remains $$m$$ everywhere.
The weight of the object at height $$h=2R$$,
$$F=\dfrac{GMm}{(h+R)^2}=\dfrac{GMm}{9R^2}=\dfrac{w}{9}$$
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)
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$$\displaystyle 2\pi \sqrt { \frac { R }{ 2g } } $$
0%
$$\displaystyle 2\pi \sqrt { \frac { R }{ 3g } } $$
0%
$$\displaystyle 2\pi \sqrt { \frac { R }{ g } } $$
0%
$$\displaystyle 2\pi \sqrt { \frac { 2R }{ g } } $$
Explanation
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=$$\dfrac{2\pi}{\omega}$$
$$=2\pi\sqrt{\dfrac{2R}{g}}$$
When a body is at a depth 'd' from the earth surface its distance from the centre of the earth is _______
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$$(R - d)$$
0%
$$2(R - d)$$
0%
$$(3R - d)$$
0%
$$(R - 2d)$$
Explanation
$$SP=d$$
$$SQ=r$$
$$PQ=(R-d)=$$ Distance of the from the centre of the Earth.
How far above the earth's surface must an astronaut in space be if they are to feel a gravitational acceleration that is half what they would feel on the surface of the earth?
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$${R}_{Earth}$$
0%
$$2{R}_{Earth}$$
0%
$$\dfrac{{R}_{Earth}}{2}$$
0%
$$\sqrt{2}{R}_{Earth}$$
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$$\sqrt{2}{R}_{Earth}-{R}_{Earth}$$
Explanation
Let the radius of earth be $$R_{Earth}$$ and acceleration due to gravity at the surface be $$g$$.
Thus acceleration due to gravity at a height h above the surface $$g' = g\dfrac{R_{Earth}^2}{(R_{Earth } + h)^2} $$
$$\therefore$$
$$\dfrac{g}{2} = g\dfrac{R_{Earth}^2}{(R_{Earth } + h)^2} $$
OR $$R_{Earth} + h = \sqrt{2} R_{Earth}$$ $$\implies h = \sqrt{2}R_{Earth} - R_{Earth}$$
Every planet revolves around the sun in a/an ______ orbit.
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elliptical
0%
circular
0%
parabolic
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none of these
Explanation
The first law of Kepler states that t
he orbit
of a planet is an
el
lipse
with the Sun at one of the two
foci.
Hence correct answer is option $$A $$
The ratio of SI unit of G to its CGS unit is _______.
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$$100 : 1$$
0%
$$1000 : 1$$
0%
$$10 : 1$$
0%
$$10000 : 1$$
Explanation
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 $$B$$
An astronaut who weighs $$162$$ pounds on the surface of the earth is orbiting the earth at a height above the surface of the earth of two earth radii ($$h = 2R$$ where R is the radius of the earth.)
How much does this astronaut weigh while in orbit at this height (With how much force is the earth pulling on him while he is in orbit at this height?)
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$$81\ pounds$$
0%
$$40.5\ pounds$$
0%
$$18\ pounds$$
0%
$$54\ pounds$$
0%
$$0\ pounds$$ (astronaut is weightless)
Explanation
Given : $$h = 2R$$
Weight of astronaut on the earth surface $$W_s = 162$$ pounds
Thus weight of
astronaut at height h $$W' = W_s \bigg( \dfrac{R}{R+h} \bigg)^2$$
$$\therefore$$
$$W' = 162 \bigg( \dfrac{R}{R+2R} \bigg)^2 =\dfrac{162}{9} = 18$$
pounds
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?
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$$72 N$$
0%
$$28 N$$
0%
$$16 N$$
0%
$$32 N$$
Explanation
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 $$D$$
State whether true or false.
The value of acceleration due to gravity becomes half at a depth of half the radius of the earth.
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True
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False
Explanation
$$g=\cfrac { GM }{ { R }^{ 2 } } $$
Let 'p' be the density of the material of earth
mass = volume * density
M=$$\cfrac { 4 }{ 3 } \pi { R }^{ 3 }p$$
Substituting in eq 1 we get
$$g=\cfrac { G }{ { R }^{ 2 } } *\cfrac { 4 }{ 3 } \pi { R }^{ 3 }p=\cfrac { 4 }{ 3 } \pi GRp$$ - eq. 2
Now at depth 'd'
$${ g }_{ d }=\cfrac { 4 }{ 3 } \pi G(R-d)p$$
or $${ g }_{ d }=g(1-\cfrac { d }{ R } )$$
if d=R/2
then $$g_{ d }=g(1-\cfrac { R }{ 2R } )=g(1-\cfrac { 1 }{ 2 } )=\cfrac { 1 }{ 2 } g$$
Thus value of acceleration due to gravity at a depth of half of the radius of the earth.
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?
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$${ P }_{ 1 }$$
0%
$${ P }_{ 4 }$$
0%
$${ P }_{ 3 }$$
0%
$${ P }_{ 2 }$$
Explanation
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 $$P_1$$
$$\textbf{Hence option A correct}$$
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 :
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$$2R$$
0%
$$R$$
0%
$$R/2$$
0%
$$R/4$$
Explanation
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,
$$W_h = \dfrac{W_s}{4}$$
$$\dfrac{GMm}{(R+h)^2} = \dfrac{GMm}{4R^2}$$
$$\dfrac{1}{R+h}=\dfrac{1}{2R}$$
$$R+h=2R$$
$$h=R$$
At what height from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be $$\dfrac {g}{2}?$$
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$$R\sqrt {2}$$
0%
$$R(\sqrt {2} - 1)$$
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$$R$$
0%
$$R(\sqrt {2} - 2)$$
Explanation
We have, $${ g }_{ h }=\dfrac { g }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$
$${ g }_{ h }=\dfrac { g }{ 2 } $$
$$\therefore \dfrac { g }{ 2 } =\dfrac { g }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$
$$\Rightarrow \dfrac { 1 }{ 2 } =\dfrac { 1 }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } $$
$${ \left( 1+\dfrac { h }{ R } \right) }^{ 2 }=2$$
$$1+\dfrac { h }{ R } =\sqrt { 2 }$$
$$\dfrac { h }{ R } =\sqrt { 2 } -1$$
$$h=R\left( \sqrt { 2 } -1 \right) $$
At what height from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be $$\dfrac {4g}{9}?$$
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$$9R/4$$
0%
$$R/2$$
0%
$$3R/2$$
0%
$$4R/9$$
Explanation
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, $$h=\cfrac { R }{ 2 } $$
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)
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$$400\ N$$
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$$3.6\ N$$
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$$1\ N$$
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$$4\ N$$
Explanation
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 $$m{ g }_{ h }=\dfrac { 400 }{ 81 } \approx 4N$$
A body weighs $$72\ N$$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?
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$$72\ N$$
0%
$$28\ N$$
0%
$$16\ N$$
0%
$$32\ N$$
Explanation
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 $$D$$
At a place, the value of 'g' is less by $$1$$% than its value on the surface of the Earth (Radius of Earth, $$R = 6400\ km$$). The place is:
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$$64\ km$$ below the surface of the earth
0%
$$64\ km$$ above the surface of the earth
0%
$$30\ km$$ above the surface of the earth
0%
$$32\ km$$ below the surface of the earth
Explanation
$${ g }_{ d }=g\left( 1-\cfrac { d }{ R } \right) $$
Now, $${ g }_{ d }=g-\cfrac { g }{ 100 } =\cfrac { 99 }{ 100 } g\\ \therefore \cfrac { 99 }{ 100 } g=g\left( 1-\cfrac { d }{ 6400\times { 10 }^{ 3 } } \right) \\ or,\quad \cfrac { 99 }{ 100 } =1-\cfrac { d }{ 64\times { 10 }^{ 5 } } \\ \Rightarrow \cfrac { 1 }{ 100 } =\cfrac { d }{ 64\times { 10 }^{ 5 } } \\ \Rightarrow d=\cfrac { 64\times { 10 }^{ 5 } }{ 100 } =64\times { 10 }^{ 3 }m=64km$$.
A spring balance is graduated when it is at sea level. If a body is weighed at consecutively increasing heights from earth's surface, the weight indicated by the balance:
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will go on increasing continuously
0%
will go on decreasing continuously
0%
will remain same
0%
will first increase and then decrease
Explanation
The acceleration due to gravity will be decreasing with increasing of height from the earth's surface. Since the weight is the products of mass and acceleration due to gravity, so the weight will decrease with increasing height from earth's surface.
As the planet revolves from point P to point Q, the velocity of the planet :
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increases
0%
decreases
0%
remains same
0%
remains equal in magnitude but opposite in direction
Explanation
When a planet resolves round the sun, the orbital velocity of the planet is $$v=\sqrt{\dfrac{GM_s}{r}}$$
where $$G=$$ gravitational constant, $$M_s=$$ mass of the sun and $$r=$$ distance from the planet to the center of the sun.
As $$G, M_s$$ are constant, so $$v\propto \dfrac{1}{\sqrt r}$$
As distance $$SP>SQ$$ , so velocity at Q is more than that at P.
Thus, velocity increases from P to Q.
At what depth from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be: $$\dfrac {2g}{5}$$?
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$$2/5R$$
0%
$$3/5R$$
0%
$$4/25R$$
0%
$$9/25R$$
Explanation
Gravity due to the earth is at depth $$g$$'=$$g(1-\dfrac{d}{R})$$,
Where $$d$$= depth from the surface of the earth, $$R$$= Radius of the earth,
Given that $$g$$'= $$\dfrac{2g}{5}$$
$$\Rightarrow \dfrac{2g}{5}=g(1-\dfrac{d}{R})$$,
Solving this $$\Rightarrow d=\dfrac {3}{5}R$$
At what height from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be $$\dfrac {g}{100}?$$
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$$10R$$
0%
$$9R$$
0%
$$100R$$
0%
$$R/100$$
Explanation
We have, $${ g }_{ h }=\dfrac { g }{ { \left( 1+\frac { h }{ R } \right) }^{ 2 } } $$
$${ g }_{ h }=\dfrac { g }{ 100 } $$
$$\therefore \quad \dfrac { g }{ 100 } =\dfrac { g }{ { \left( 1+\cfrac { h }{ R } \right) }^{ 2 } } \Rightarrow \dfrac { 1 }{ 100 } =\dfrac { 1 }{ { \left( 1+\cfrac { h }{ R } \right) }^{ 2 } }$$
$$100={ \left( 1+\cfrac { h }{ R } \right) }^{ 2 }\Rightarrow 10=1+\cfrac { h }{ R }$$
or, $$h=9R$$
As the altitude increases, the acceleration due to gravity:
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Remains constant
0%
Becomes zero
0%
Decreases
0%
Increases
Explanation
$$g=\frac{Gm}{R^{2}}$$
Let acceleration due to gravity at a height $$h$$ =$$g_{1}$$
$$g_{1}=\frac{Gm}{(R+h)^{2}}$$
$$\frac{g_{1}}{g}=(1+\frac{h}{R})^{-2}$$
Expanding by binomial theorem and neglecting higher powers
$$\frac{g_{1}}{g}=(1-2\frac{h}{R})$$
$$g_{1}=g(1-2\frac{h}{R})$$
So as altitude increases acceleration decreases
The value of universal gravitational constant $$G$$ is-
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$$6.67\times 10^{-11} \dfrac{Nm^2}{kg}$$
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$$6.67\times 10^{-11} \dfrac{Nm^2}{kg^2}$$
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$$66.7\times 10^{-11} \dfrac{Nm^2}{kg^2}$$
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$$66.7\times 10^{-11} \dfrac{Nm^2}{kg}$$
Explanation
Gravitational force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance $$r$$, $$F = \dfrac{Gm_1 m_2}{r^2}$$
$$\implies$$ $$G = \dfrac{F r^2 }{m_1 m_2}$$
Value of universal gravitational constant $$G$$ is $$6.67\times 10^{-11}$$ $$\dfrac{Nm^2}{kg^2}$$.
S.I. Unit of universal gravitational constant $$G$$ is-
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$$\dfrac{Nm^2}{Kg}$$
0%
$$\dfrac{Nm^2}{Kg^2}$$
0%
$$\dfrac{Nm}{Kg^2}$$
0%
$$\dfrac{Nm}{Kg}$$
Explanation
Gravitational force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance $$r$$, $$F = \dfrac{Gm_1 m_2}{r^2}$$
$$\implies$$ $$G = \dfrac{F r^2 }{m_1 m_2}$$
SI unit of universal gravitational constant $$G$$ is $$\dfrac{Nm^2}{Kg^2}$$.
If the radius of the earth shrinks by $$1.5$$% (mass remaining same), then the value of acceleration due to gravity changes by :
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$$1$$%
0%
$$2$$%
0%
$$3$$%
0%
$$4$$%
Explanation
Acceleration due to gravity on the earth's surface is $$g=\dfrac{GM}{R^2}$$
When the earth shrinks $$1.5$$ %, the radius becomes $$R'=R-R(1.5/100)=(9.85/100)R$$ or $$R'/R=98.5/100$$
After shrinks, the acceleration due to gravity becomes, $$g'=\dfrac{GM}{R'^2}$$
So, $$\dfrac{g'}{g}=(R/R')^2=(100/98.5)^2=1.03$$
Thus, percentage change in acceleration $$=\dfrac{g'-g}{g}\times 100=[g'/g -1]100=[1.03-1]100=3$$ %
The value of g on the earth's surface is $$980\ cm\ s^{-2}$$. Its value at a height of $$64\ km$$ from the earth's surface is
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$$960.40\ cm\ s^{-2}$$
0%
$$984.90\ cm\ s^{-2}$$
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$$982.45\ cm\ s^{-2}$$
0%
$$977.55\ cm\ s^{-2}$$
Explanation
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, $$g_h=\dfrac{gR^2}{(R+h)^2}=\dfrac{g}{(1+h/R)^2}=\dfrac{980}{(1+64/6400)^2}=960.69 cm s^{-2}$$
Variation of acceleration due to gravity $$\left(g\right)$$ with distance $$x$$ from the centre of the earth is best represented by : ($$R \rightarrow$$ Radius of the earth)
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0%
0%
0%
Explanation
Acceleration due to gravity at a depth $$d$$ from the earth surface $$g = g_o\bigg(1-\dfrac{d}{R}\bigg)$$
Thus distance from earth's centre $$x = R-d$$ i.e. $$d = R-x$$
$$\implies$$
$$g = g_o\dfrac{x}{R}$$
Thus a
cceleration due to gravity increases linearly with the increase in distance from centre of earth.
Acceleration due to gravity at a height $$h$$ from the earth surface $$g = g_o\bigg(\dfrac{R}{R+h} \bigg)^2$$
Thus distance from earth's centre $$x = R+h$$ i.e. $$h = x-R$$
$$\implies$$
$$g = g_o\bigg(\dfrac{R}{x} \bigg)^2$$
Thus a
cceleration due to gravity decreases as $$\dfrac{1}{x^2}$$ with the increase in distance from centre of earth.
Hence, option D is correct.
At the centre of the earth acceleration due to gravity is:
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0%
zero
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infinity
0%
9.8
0%
98
Explanation
gravitational field $$g$$ at distance $$r$$ where $$r<R$$( radius of earth) is
$$g=\frac{G\rho r}{3}$$ ($$\rho$$ is density of earth)
So at centre $$r=0$$
hence $$g=0$$
So at the centre of earth acceleration due to gravity is zero
At a point very near to the earth's surface, the acceleration due to gravity is g. What will be the acceleration due to gravity at the same point if the earth suddenly shrinks to half its radius without any change in its mass?
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$$2\ g$$
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$$4\ g$$
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$$g$$
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$$3\ g$$
Explanation
Acceleration due to gravity on the earth's surface is $$g=\dfrac{GM}{R^2}$$
If earth suddenly shrinks to half its radius, the acceleration due to gravity becomes, $$g'=\dfrac{GM}{(R/2)^2}=4GM/R^2=4g$$
Dimensional formula of universal gravitational constant $$G$$ is-
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$$M^{-1}L^3T^{-2}$$
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$$M^{-1}L^2T^{-2}$$
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$$M^{-2}L^3T^{-2}$$
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$$M^{-2}L^2T^{-2}$$
The minimum speed of a particle projected from earth's surface so that it will never return is/are:
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$$\dfrac{GM}{R}$$
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$$22.1 km/s$$
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$$(4g_oR)$$
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none of above
Explanation
An object can be thrown up with a certain minimum initial velocity so that, the object goes beyond the earth's gravitational field and escape from earth, this velocity
escape velocity of the earth.
potential energy at surface is $$-\frac{GMm}{R}$$, so kinetic energy that has to give to reach particle to infinite ( zero energy) is equal to potential energy.
$$\frac{1}{2}mv_e^2=\frac{GMm}{R}=\sqrt{\frac{2GM}{R}}$$.
so none ofthe above give this expression so the answer is D.
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:
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$$11.1\ J$$
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$$5.5\ J$$
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$$16\ J$$
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$$27.7\ J$$
Explanation
Let acceleration due to gravity in the planet is $$g_p$$
Then gravitational potential difference
=$$g_ph$$
$$16=g_p\times 20$$
$$g_p=\dfrac{16}{20}=\dfrac{4}{5}m/s^2$$
Workdone=$$mg_ph=2\times \dfrac{4}{5}\times8\times \sin60$$
$$\implies 11.09J$$
A particle hanging from a massless spring stretches it by $$2\ cm$$ at earth's surface. How much will the same particle stretch the spring at height $$2624\ km$$ from the surface of earth? ( Radius of earth $$=6400\ km$$)
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$$1\ cm$$
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$$2\ cm$$
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$$3\ cm$$
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$$4\ cm$$
Explanation
$$x=2cm=0.02m$$
$$h=2624km$$
$$R=6400km$$
$$\therefore$$ $$\cfrac { { x }^{ \prime } }{ x } =\cfrac { { g }^{ \prime } }{ g } =\cfrac { { 6400 }^{ 2 } }{ { \left( 6400+2624 \right) }^{ 2 } } \approx 0.5\\ { x }^{ \prime }=0.5\times x=0.5\times 2cm=1cm$$
If earth's radius were to hypothetically shrink by $$1\%$$, the value of $$G$$ would:
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Shrink by $$1\%$$
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Expand by $$1\%$$
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Remain the same
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Shrink by $$0.01\%$$
Explanation
According to the universal law of gravitation, every object in the universe attracts every other objects with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects.
$$F=G\dfrac{m_1m_2}{d^2}$$
where $$G$$ is the constant of proportionality and
is called the universal gravitation constant. It does not depend on the radius.
So, if the earth's radius shrinks by $$1\%$$, the value of $$G$$ would remain the same.
The SI unit of gravitational potential is
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$$J$$
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$$Jkg^{-1}$$
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$$Jkg$$
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$$Jkg^{-2}$$
Explanation
It is a very simple idea. Gravitational potential is the potential energy per kilogram at a point in a field. So the units are joule
per kilogram.
so the best possible answer is option B.
How the gravitational constant will change if a brass plate is introduced between two bodies?
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No change
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Decreases
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Increases
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No sufficient data
Explanation
Universal Gravitational Constant (G) remains constant irrespective of the medium between two bodies. It has a constant value of $$6.67 \times 10^{-11} Nm^{2}kg^{-2}$$
The minimum energy required to launch a $$m$$ kg satellite from earth's surface in a circular orbit at an altitude of $$2R$$ where $$R$$ is the radius of earth, will be:
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$$3mgR$$
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$$\dfrac{5}{6} mgR$$
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$$2mgR$$
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$$\dfrac{1}{5} mgR$$
Explanation
The kinetic energy at altitude $$2R$$ is $$= \dfrac{G M m} {6 R}$$
The gravitational potential energy at altitude $$2R$$ is $$=\dfrac{– G M m}{ 3 R}$$
Total energy $$=$$ Kinetic energy $$+$$ potential energy
$$= \dfrac{G M m} {6 R}+\dfrac{– G M m}{ 3 R}$$
Potential energy at the surfce is
$$= \dfrac{G M m} { R}$$
Therefore, required kinetic energy $$= -\dfrac{G M m} {6 R}+ \dfrac{G M m} { R}= \dfrac{5G M m} { 6R}$$
hence the best possible answer is option B.
Let $$V$$ and $$E$$ be the gravitational potential and gravitational field at a distance $$r$$ from the centre of a uniform spherical shell. Consider the following two statements, ($$A$$) The plot of $$V$$ against $$r$$ is discontinuous and ($$B$$) The plot of $$E$$ against $$r$$ is discontinuous.
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Both A and B are correct
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A is correct but B is wrong
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B is correct but A is wrong
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both A and B are wrong.
Explanation
Let $$V$$ be the gravitational potential and $$E$$ be the gravitational field at a distance $$r$$ from the center of the sphere.
Graphically as shown in above images.
From IMAGE $$1$$ and IMAGE $$2$$
IMAGE $$1$$ Plot of $$V$$ against $$r$$ is continuous
.
IMAGE $$2$$ Plot of $$E$$ against $$r$$ is discontinuous.
Thus condition $$A$$ is wrong and $$B$$ is correct.
If a particle is slowly brought from reference point to another point $$P$$ in a gravitational field, then work done per unit mass by the external agent is (at that point)
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gravitational force
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gravitational field intensity
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gravitation potential
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none of the above
Explanation
If a particle is slowly brought from reference point to another point P in a gravitational field, then work done per unit mass by external agent is (at that point)
gravitational potential.
A body of mass 'm' is approaching towards the centre of a hypothetical hollow planet of mass 'M' and radius 'R'. The speed of the body when it passes the centre of the planet through a diametrical tunnel is:
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$$\sqrt {\dfrac{GM}{R}}$$
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$$\sqrt {\dfrac{2GM}{R}}$$
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Zero
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none of these.
The value of g at a particular point is 9.8 $$m/sec^2$$ suppose the earth suddenly shrink uniformly to half its present size without losing any mass. The value of $$g$$ at the same point (assuming that the distance of the point from the centre of the earth does not shrink) will become
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$$9.8m/sec^2$$
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$$4.9m/sec^2$$
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$$19.6m/sec^2$$
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$$3.1m/sec^2$$
Explanation
$$g=\cfrac { GM }{ r } $$
Here $$M$$ and $$r$$ are constant
$$\therefore { g }^{ \prime }=9.8m{ s }^{ -2 }$$
The energy required to remove a body of mass $$m$$ from earth's surface is/are equal to:
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$$\dfrac{-GMm}{R}$$
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$$mgR$$
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$$-mgR$$
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none of these.
Explanation
An object can be thrown up with a certain minimum initial velocity so that, the object goes beyond the earth's gravitational field and escape from earth, this velocity
escape velocity of the earth.
potential energy at surface is $$-\frac{GMm}{R}$$, so kinetic energy that has to give to reach particle to infinite ( zero energy) is equal to potential energy.
$$E=\frac{GMm}{R}=g.m.R$$
so the best possible answer is option B.
A hypothetical planet has density $$\rho$$, radius R, and surface gravitational acceleration g. If the radius of the planet were doubled, but the planetary density stayed the same, the acceleration due to gravity at the planet's surface would be.
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$$4g$$
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$$2g$$
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g
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$$g/2$$
Explanation
The minimum speed of a particle projected from earths surface so that it will never return is/are:
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(GM/R)
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22.1 km/sec
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$$(4g_oR)$$
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none of above
Explanation
The minimum speed of a particle projected from earth's surface so, that it will never return is escaped velocity $$=\sqrt{\cfrac{2GM}{R}}$$
The value of acceleration due to gravity at a depth of $$1600 km$$ is equal to [Radius of earth $$= 6400 km$$]
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$$9.8{ ms }^{ -2 }$$
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$$4.9{ ms }^{ -2 }$$
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$$7.35{ ms }^{ -2 }$$
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$$19.6{ ms }^{ -2 }$$
Explanation
Given : $$d = 1600 km$$ $$R = 6400 km$$
Acceleration due to gravity at a depth $$d$$, $$g_d = g_s\bigg(1 -\dfrac{d}{R}\bigg)$$
where $$g_s = 9.8 ms^{-2}$$ is acceleration due to gravity at earth's surface.
$$\therefore$$
$$g_d = 9.8 \bigg(1 -\dfrac{1600}{6400}\bigg) = 7.35 ms^{-2}$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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