Decreases with height from earth's surface

Increases with height from earth's surface

Remain unchanged with increase of height from earth's surface

A group of small heavenly bodies, forming a belt rotating round the sun

Starts forming a recognizable pattern

Solid bodies from the outer space

A large cluster of stars distributed in the universe in an irregular form

MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 2

Acceleration due to gravity ---------- with height from the surface of the earth.

0%
Decreases
0%
Increases
0%
Remains constant
0%
Cannot say

Explanation

Any two objects attract each other with a force

$F_g=\dfrac{GMm}{R^2}$ where

$G$ is the gravitational constant,

$M$ and

$m$ the masses of the two bodies and

$R$ is the distance along the line joining their centre of mass

Escape velocity of a particle depends on its mass m as

0%
$m^2$
0%
$m$
0%
$m^0$
0%
$m^{_1}$

Explanation

Escape velocity,

$v_e=\sqrt{2gR}$
it is independent of the mass of the particle.
Thus it will depend on

$m^0$ .
hence, option C is correct.

SI unit of G is

$Nm^{2}kg^{-2}$ .Which of the following can also be used as the SI unit of G?

0%
$m^3 kg^{-1} s^{-2}$
0%
$m^2 kg^{-2} s^{-1}$
0%
$m kg^{-1} s^{-1}$
0%
$m^3kg^{-3}s^{-2}$

Explanation

Here

$N$ represents Newton.

so we can write unit of force in place of newton, also

$F=ma$

so we can write

$G=M^1L^1t^{-2}L^2M^{-2}=M^{-1}L^3T^{-2}$

SI unit for mass is kilogram(kg), for length is meter(m), for time is sec(s)

so we can write unit for G in SI as

$kg^{-1}m^3s^{-2}$

so best possible answer is option A.

The value of gravitational acceleration g at the center of the earth is:-

0%
Infinite
0%
$\displaystyle 9.8{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 32.2{ m }/{ { s }^{ 2 } }$
0%
Zero

Explanation

acceleration due to gravity varies directly proportional to distance from center, i.e. r<R.

But varies

$\dfrac{1}{r^2}$ when r>R

expression is given by

$g=\dfrac{GM}{r^2}$ if r>R.

and

$g=\dfrac{GMr}{R^3}$ if r<R

so when

$r=0$ then g will be zero.

The value of '

$g$ ' is zero

0%
At the top atmosphere
0%
At $20 km$ below the surface of the earth
0%
At $20 km$ above the surface of the earth
0%
At the centre of the earth

The ratio of the value of G in SI units to CGS units is

0%
$10^3:1$
0%
$10^2:1$
0%
$10^{-2}:1$
0%
$10^{-3}:1$

Explanation

$\text{Quantity}$	$\text{SI System}$	$\text{CGS System}$
Force	Newton (N)	dyne
Distance	metre (m)	centi-meter (cm)
Mass	kilogram (kg)	gram (g)

The unit of G can be calculated by putting the unit of all the quantities in the expression of the force due to gravity.

$\dfrac{Gm_1m_2}{r^2}=F$

$\Rightarrow G=\dfrac{F\times r^2}{m_1\times m_2}$

So we get,

$\text{Unit of G}=\dfrac{\text{Unit of force}\times \text{(Unit of distance)}^2}{\text{(Unit of mass)}^2}$

$\text{SI unit of G}=\dfrac{N.m^2}{kg^2}$ and

$\text{CGS unit of G}=\dfrac{dyne.cm^2}{g^2}$

We know that

$1\ N = 10^5\ dyne$ ,

$1\ kg = 10^3\ g$ and

$1\ m = 10^2\ cm$

$\underline{\text{Required ratio:}}$

$\dfrac{SI}{CGS}=\dfrac{N.m^2.g^{-2}}{dyne.cm^2.kg^{-2}}=\dfrac{10^{5}\ dyne\times 10^4\ cm^2\times g^{-2}}{dyne\times cm^2\times 10^{-6}\ g^{-2}}=10^3 : 1$

So the answer is option A.

The value of G depends on

0%
the nature of the interacting bodies
0%
the size of the interacting bodies
0%
the mass of the interacting bodies
0%
none of these

Where will a body weigh minimum?

0%
At a height of $100 m$ above the earth's surface
0%
At the earth's surface
0%
At a depth of $100 m$ below the earth's surface
0%
At the centre of the earth

Explanation

Weight at center of earth is 0.

$g=\dfrac{4\pi}{3} {G\rho (R-d)}$
Thus, for

$d=R$ i.e. at centre of earth

$g=0$ and hence

$W=0$

The value of G does not depend on

0%
nature of the interacting bodies
0%
size of the interacting bodies
0%
mass of the interacting bodies
0%
all the above

Explanation

G or the Universal gravitational constant remains constant throughout the universe and its value is

$6.67 \times 10^{-11} m^3kg^{-1} s^{-2}$ . It is not affected by any factor.

$F = \dfrac{GmM}{r^2} \ \Rightarrow F \propto \dfrac{Mm}{r^2}$

G is a proportionality constant and is independent of all variables.

The value of

$g$

0%
Decreases with height from earth's surface
0%
Increases with height from earth's surface
0%
Remain unchanged with increase of height from earth's surface
0%
None of these

Explanation

$g_h=g(1-\cfrac{2h}{r})$
So, The value of acceleration due to gravity decreases with increase in height above the surface of the earth.
hence,option A is correct.

The three laws of planetary motion were given by:

0%
Einstein
0%
Kepler
0%
Copernicus
0%
Tycho Brahe

Asteroids are

0%
A group of small heavenly bodies, forming a belt rotating round the sun
0%
Starts forming a recognizable pattern
0%
Solid bodies from the outer space
0%
A large cluster of stars distributed in the universe in an irregular form

The maximum value of

$g$ is

0%
At the poles
0%
At the top of the Mounts Everest
0%
At the equator
0%
Below the sea level

Explanation

The value of

$'g'$ depends upon the distance from the centre of the earth; lesser thea distance higher the value. At the poles this distance is least, hence

$g$ is maximum there.

A body weighs 60 kg on the earth's surface. What would be its weight at the centre of the earth?

0%
60 kg-wt
0%
6 kg-wt
0%
60 $\times$ 9.8 kg-wt
0%
zero

A raised hammer possesses

0%
K.E. only
0%
Gravitational P.E,
0%
Electrical energy
0%
sound energy

Explanation

The energy possesed by a body due to virtue of its position is called potential energy.

A raised hammer will not have any velocity so it won't have any kinetic energy however it will posses potential energy due to the work done on the hammer by the gravity.

Hence correct answer is option

$B$

The variation of g with height or depth (r) is shown correctly by which of the following graphs. (where R is the radius of the earth).

Explanation

Variation of g with height (r)

$=\dfrac{GM}{(R+r) ^2}$

$\Rightarrow g$

$\alpha$

$r^{-2}$

Variation of g with depth (r)

$=\dfrac{GM}{R ^2} (1- \dfrac{r}{R})$

$\Rightarrow g$

$\alpha$

$r$

Hence correct answer is option

$A$

The evidence to show that there must be force acting on Earth and directed towards the Sun is.

0%
phenomenon of day and night
0%
apparent motion of the Sun around the Earth
0%
revolution of Earth around the Sun
0%
deviation of the falling bodies towards east

Which of the following units can be used to express G?

0%
N Kg $^{-2}$
0%
Nm $^{-2}$ Kg $^{2}$
0%
Nm $^{2}$ Kg $^{-2}$
0%
Nm $^{-2}$ Kg $^{-3}$

Explanation

Gravitational force between two masses

$m$ and

$M$ , separated by a distance

$r$ is given by

$F = \dfrac{GMm}{r^2}$

Thus,

$G = \dfrac{Fr^{2}}{Mm}$

We know that the unit of force (

$F$ ) is newton (N), the unit of mass (

$m\ or\ M$ ) is kg and the unit of distance (

$r$ ) is meter(m).

$\text{The unit of G} = \dfrac{Nm^2}{kg^2} = Nm^2kg^{-2}$

In a vacuum on the earth surface, all freely falling bodies

0%
have the same speed
0%
have the same velocity
0%
have the same force
0%
have the same acceleration

Explanation

a force acting on the body is given by

$F=\dfrac{GMm}{r^2}$

here

$m$ is different for different bodies hence force acting on different bodies will be different.

acceleration is

$\dfrac{force}{mass}=\dfrac{GM}{r^2}$ , this is independent of the mass of the body or type of the body.

also, speed or velocity depends on time for which it is accelerated because acceleration is the same.

hence correct option is D.

At a height equal to earth's radius, above the earth's surface, the acceleration due to gravity is

0%
g
0%
$\displaystyle \frac{g}{2}$
0%
$\displaystyle \frac{g}{4}$
0%
$\displaystyle \frac{g}{8}$

Explanation

Acceleration due to gravity =

$g = \dfrac{GM_{earth}}{R^2}$ ,

$R$ = distance from the centre of the earth.

At a height = R above earth's surface

$g = \dfrac{GM_{earth}}{(R+R)^2} = \dfrac{GM_{earth}}{4R^2} = \dfrac{g}{4}$ Option C.

At the centre of the earth, the value of g becomes

0%
zero
0%
unity
0%
infinity
0%
none of these

Explanation

acceleration due to gravity varies directly proportional to distance from center, i.e. r<R.

But varies

$\dfrac{1}{r^2}$ when r>R

expression is given by

$g=\dfrac{GM}{r^2}$ if r>R.

and

$g=\dfrac{GMr}{R^3}$ if r<R

so when

$r=0$ then g will be zero.

so the answer is option A.

State whether true or false.

As the distance of the planet from the sun increases, the period of revolution decreases.

0%
True
0%
False

Calculate the weight of the block at a distance of four times of Earth radii from the center of the Earth. It is given that g is the acceleration due to gravity at the surface of Earth and mass of the block is m.

0%
16mg
0%
4mg
0%
2mg
0%
$\dfrac{1}{4}$ mg
0%
$\dfrac{1}{16}$ mg

Explanation

Acceleration due to gravity at earth's surface

$g = \dfrac{GM_{earth}}{R^2}$ ,

$R$ = distance from the centre of the earth.

At a height

$h= 4R$ from earth's center

$g' = \dfrac{GM_{earth}}{(4R)^2} = \dfrac{GM_{earth}}{16R^2} = \dfrac{g}{16}$

So, weight

$W' = mg' = \dfrac{mg}{16}$

Fill in the blanks:

Value of gravitational constant (G) on moon is __________ as on earth.

0%
Greater
0%
Smaller
0%
Same
0%
None

Kepler's laws of motion are applicable to _______

0%
All objects
0%
Planets
0%
Objects on earth
0%
Objects on moon

If the radius of the earth were to shrink by

$2$ %, its mass remaining the same, the acceleration due to gravity on the earth's surface would

0%
decrease by $2$ %
0%
increase by $2$ %
0%
increase by $4$ %
0%
decrease by $4$ %

Explanation

$g=\cfrac { GM }{ { R }^{ 2 } }$ , if

$R$ decreases, then

$g$ increases
Taking log on both sides, we have

$\log { g } =\log { G } +\log { M } -2\log { R }$
Differentiating it we get

$\cfrac { dg }{ g } =0+0-\cfrac { 2dR }{ R }$

$=-2\times \left( \cfrac { -2 }{ 100 } \right) =\cfrac { 4 }{ 100 }$

$=\cfrac { 4 }{ 100 } \times 100$

$=4$ %

When an object is far from earth, 'g' is calculated by the formula:

0%
$G\frac{M\times m}{R^2}$
0%
$G\frac{M}{R^2}$
0%
$G\frac{M\times m}{d^2}$
0%
$G\frac{M}{d^2}$

Explanation

When an object in on or near the surface of the earth, the expression for

$g$ may be approximated as:

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

But, for objects far away from the surface of the earth,

$g \propto \dfrac{1}{d^2}$

Hence

$g=\dfrac{GM}{d^2}$ would be the correct expression.

When an object is close to the surface of earth the acceleration due to gravity is calculated by using the formula

0%
$\frac{F\times d^2}{M\times m}$
0%
$G\frac{M\times m}{d^2}$
0%
$G\frac{M}{R^2}$
0%
$G\frac{M}{d^2}$

Explanation

When an object is close to the surface of earth, the acceleration created due to gravity is constant and calculate by the formula

$g=G\frac{M}{R^2}$
Where, G= gravitation constant
M = Mass of earth
R = Radius of earth
Since the distance for object close to the earth will not change the value of radius of earth in large magnitude, it is considered to calculate acceleration due to gravity for objects close to earth.

The escape velocity nearer to the surface of a planet of radius

$6800$ km and mass

$\displaystyle 8 \times {10} ^{24}$ kg is ____

$\displaystyle {kms} ^{-1}$

0%
$32.5$
0%
$21.2$
0%
$12.5$
0%
$11.2$

Explanation

Escape velocity of a satellite nearer to the surface of a planet is given by

Escape velocity =

$\sqrt{\dfrac{2GM}{R}}$

$\sqrt{\dfrac{2 \ast 9.8 \ast 8 \ast 10^{24}}{6800}}$

= 12.5 Kms

$^{-1}$

Universal gravitational constant

$G$ is proportional to distance

$r$ between two masses as-

0%
$r^2$
0%
$r$
0%
$1/r$
0%
does not depend upon $r$

Explanation

According to Newton's Law of gravitation, the gravitational force between two bodies of mass

$M$ and

$m$ separated by a distance

$r$ , is directly proportional to the product of the masses of the objects (

$F \propto Mm$ ) and inversely proportional to the square of distance between them

$F \propto \dfrac{1}{r^2}$ .

When we combine this information, we get

$F \propto \dfrac{Mm}{r^2}$

$F= G \dfrac{Mm}{r^2}$

where

$G$ is the proportionality constant, called the Universal Gravitational constant.

Since

$G$ is a constant it does not depend on

$r$ .

Value of universal gravitational constant

$G$ in CGS unit is-

0%
$6.67\times 10^{8}cm^3g^{1}s{}$
0%
$6.67\times 10^{8}cm^3g^{-1}s^{-2}$
0%
$6.67\times 10^{9}cm^3g^{1}s^{2}$
0%
$6.67\times 10^{7}cm^3g^{-1}s^{2}$

Explanation

Newton's law of gravitation:

$F=G \dfrac{Mm}{r^2}$

$G = \dfrac{F r^2}{Mm}$

In CGS unit system:

r - distance - cm,

M,m - mass - grams,

F - force - dyne -

$g cm/s^2$

Therefore units of G -

$\dfrac{(g \space cm/s^2)(cm^2)}{g^2} = cm^3g^{-1}s^{-2}$

The universal law of gravitation must be applicable to

0%
The earth and the moon.
0%
The planets around the Sun.
0%
Any pair of bodies.
0%
The earth and the apple.

Value of acceleration due to gravity of earth is maximum :

0%
At centre of earth
0%
At surface of earth
0%
At a height of 50 km from earths surface
0%
At a height of 12 km from earths surface

Explanation

Acceleration due to gravity at a height

$h$ above the surface

$g' = g\bigg( 1-\dfrac{2h}{R}\bigg)$

This suggest that acceleration due to gravity decreases as we go up higher above the earth surface.

At the surface, we have

$h = 0$

Thus acceleration due to gravity at the surface

$g' = g$ (maximum).

Also acceleration due to gravity is zero at the centre of earth.

Thus acceleration due to gravity is maximum at the earth's surface.

The ratio of the value of gravitational constant

$G$ between Earth and Moon system and Earth and Sun system is:

0%
Greater than $1$
0%
Less than $1$
0%
Equal to $1$
0%
Can't be determined

Explanation

According to Newton's law of gravitation, every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$F=\propto \dfrac{Mm}{r^2}$

$F=G \dfrac{Mm}{r^2}$

Gravitational constant

$G$ is the proportionality constant which has fixed value

$6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$ .

Therefore, the ratio of the value of gravitational constant

$G$ between Earth and Moon system and Earth and Sun system is equal to 1. (Because it has the same value)

The energy required to remove a body of mass 'm' from earths surface is/are equal to:

0%
-GMm/R
0%
mgR
0%
-mgR
0%
none of these.

Explanation

Potential energy required to remove a body of mass

$m$ from earth's surface is

$mgR$

State whether true or false.

The weight of a freely falling body from a very large height is always constant.

0%
True
0%
False

Explanation

Weight of a body

$w=mg$

Now

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

As a body is freely falling, its r decreases and in turn g increases.

So weight increases for a freely falling body and isn't constant.

The escape velocity of projection from the earth is approximately (

$R = 6400 km$ ).

0%
$7 km/s$
0%
$112 km/s$
0%
$11.2 km/s$
0%
$1.1 km/s$

Explanation

$V_{escape}=\sqrt{\cfrac{2GM}{R}}$

$=\sqrt{\cfrac{2(6.67\times 10^{-11}\times 5.98\times 10^{24})}{6.38\times 10^{6}}}ms^{-1}$

$=\sqrt{1.251\times 10^8} ms^{-1}$

$=11184 ms^{-1}$

$\approx 11.2 Kms^{-1}$

Motion of planets in solar system is an example of conservation of :

0%
Momentum
0%
Velocity
0%
Angular momentum
0%
Energy

Explanation

$\textbf{Explanation:}$

$\bullet$ Kepler's second law or the law of equal areas says that the line joining the planet and sun sweeps the same area in equal intervals of time. This is to conserve angular momentum.

$\bullet$ In the motion of planets around the sun the angular momentum is conserved. That is the total angular momentum of the system is constant. This is because there is no external torque acting on the system.

$\textbf{Hence option C correct}$

The earth retains its atmosphere because

0%
The earth is sphere
0%
The earth has population
0%
The escape velocity of molecules is less than the mean speed of molecules
0%
The escape velocity of molecules is more than the mean speed of molecules

The value of gravitational acceleration 'g' at a height 'h' above the earth's surface is

$\dfrac{g}{4}$ then (R = radius of earth)

0%
$h = R$
0%
$h = \dfrac{R}{2}$
0%
$h = \dfrac{R}{3}$
0%
$h = \dfrac{R}{4}$

Explanation

Value of gravitational acceleration at a height

$h$ above th earth's surface is given by

$g' = g\bigg(\dfrac{R}{R+h}\bigg)^2$

$\therefore$

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

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

$h+R = 2R$

$\implies$

$h = R$

The gravitational potential is a

0%
scalar
0%
vector
0%
scalar based on the mass of the particle
0%
scalar or vector depending on the situation

Gravitational potential energy is negative. This implies

0%
Energy is rising along negative x axis
0%
A particle is trapped in this potential
0%
The particle is moving in the opposite direction
0%
Energy is decreasing in the direction of motion of the particle

The value of

$'g'$ will be

$1$ % of its value at the surface of the earth at a height of

$(R_e = 6400km)$ -

0%
6400 km
0%
57600 km
0%
12560 km
0%
64000 km

Explanation

${g^1} = g\left( {1 - {h \over R}} \right)$

$g = g\left( {1 - {h \over R}} \right)$

$1 = {h \over R},h = R$

$R$ is the radius of the earth, then the value of acceleration due to gravity at a height

$h$ from the surface of the earth will become half its value on the surface of the earth if

0%
$h = 2R$
0%
$h = R$
0%
$h = (\sqrt {2} + 1)R$
0%
$h = (\sqrt {2} - 1)R$

Explanation

$g^{1} = \dfrac{Gm}{(R+h)^{2}} ...(1)$ & at surface

$g = \dfrac{Gm}{R^{2}} ...(2)$

given that

$g^{1} = g/2$

(1)/(2) we get

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

Put

$g^{1} = g/2$ we get

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

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

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

1187320_1106057_ans_939d6055154a473daf437a6394330aad.jpg

The weight of a body of mass

$3kg$ at a height of

$12.8 \times 10^6m$ from the surface of the earth is _______.

0%
$9.75N$
0%
$1.46N$
0%
$3.26N$
0%
$4.36N$

Explanation

Given,

$Mass=3kg$

$,h=128\times10^8=12.8\times10^7 \ m$

$R=6.4\times10^6 \ m$

So,

$h = 12.8\times10^6=2R$

$g'=g_m(\dfrac{R}{R+h})^2=9.8(\dfrac{R}{R+2R})^2=9.8(\dfrac{1}{3})^2=\dfrac{9.8}{9}$

Thus the weight,

$W=mg'=3\times \dfrac{9.8}{9}=3.26N$

Two identical spherical masses are kept at some distance. Potential energy when a mass

$m$ is taken from the surface of one sphere to the other

0%
increases continuously
0%
decreases continuously
0%
first increases, then decreases
0%
first decreases, then increases

What is the percentage decrease in the weight of a body when it is taken 32 km below the surface of the earth? [Radius of the earth=6400km]

0%
1%
0%
0.75 %
0%
0.5 %
0%
0.25 %

The value of

$'g'$ at a depth of

$80 km$ will be (Radius of earth

$= 6400 km$ and value of

$'g'$ on the surface of earth is

$10m/s^2$ )

0%
$990 \ cm/s^2$
0%
$980 \ cm/s^2$
0%
$970 \ cm/s^2$
0%
$1000 \ cm/s^2$

Explanation

Let

$g'/ge$ the acceleration due to gravity in mins

Then

$g'=g\left(1\times \cfrac {d}{R}\right)=9.8\left(1-\cfrac {640}{6400}\times 10^3\right)$

$=9.8 \times 0.9999m/s^2$

$=9.799m/s^2$

The weight of an astronaut in outer space is

0%
Less than his actual weight
0%
more than his actual weight
0%
same as his actual weight
0%
zero

If the distance between the earth and sun were to be doubled from its present value, the number of days in a year would be :

0%
64.5
0%
1032
0%
182.5
0%
730

Explanation

Given,

Radius of rotation become double, ${{a}_{2}}=2{{a}_{1}}$

By Kepler’s law, $T ^ { 2 }$ $\alpha \,\,{{a}^{3}}$

$\dfrac { T _ { 1 } ^ { 2 } } { a _ { 1 } ^ { 3 } }$ $= \dfrac { T _ { 2 } ^ { 2 } } { a _ { 2 } ^ { 3 } }$

$T _ { 2 } ^ { 2 }$ $= \left( \dfrac { a _ { 2 } } { a _ { 1 } } \right) ^ { 3 } \times T _ { 1 } ^ { 2 }$

$T _ { 2 } ^ { 2 }$ $={{\left( \dfrac{2}{1} \right)}^{3}}\times T_{1}^{2}$

$T _ { 2 } ^ { 2 }$ $=8\times T_{1}^{2}$

$T _ { 2 } ^ { 2 }$ $=\sqrt{8}\times 365\text{ days }$

${{T}_{2}}=1032.37\text{ days }$

Hence, $1032$ days in year.

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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