MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 11

Calculate the ratio of weights of a body when it is (radius of the Earth is 6400km.)

1. 200 km above the surface of the Earth and

2. 200 km below the surface of the Earth.

0%
0.97
0%
1.2
0%
2
0%
0

Explanation

The acceleration Due to Gravity variation on :-

at an height 'h'

$g_{1}^{1} = g(\dfrac{R}{R+h})^{2}$ at adept 'd'

$g_{2}^{1} = g(1-\dfrac{d}{R})$

where g : at surface and R : Radius of Earth

an weight of body, w = mg', mg',

$g'$ accln. due to gravity at that point.

Ratio of weights :-

$\dfrac{w_{1}}{w_{2}} = (\dfrac{R}{R+h})^{2}\times (\dfrac{R}{R-d})$

From question, R = 6400km, d = h = 200km so,

$\dfrac{w_{1}}{w_{2}} = \dfrac{(64000)^{3}}{(6400+200)^{2}(6400-200)} = \dfrac{(6400)^{3}}{(6600)^{2}(6200)} \simeq 0.97$ (Ans)

1101144_1172999_ans_9bbf59a90d15461485d767b585720479.JPG

If the change in the value of g at a height h above the surface of the earth is same as at depth below the surface of the earth, then (h >>R)

0%
$x = h$
0%
$x = 2h$
0%
$x = h/2$
0%
$x = h^2$

A hole is drilled from the surface of earth to its centre. A particle is dropped from rest at the surface of earth. The speed of the particle when it reaches the centre of the earth in terms of its escape velocity on the surface of earth

$v_{e}$ is :

0%
$\dfrac {v_{e}}{2}$
0%
$v_{e}$
0%
$\sqrt {2}v_{e}$
0%
$\dfrac {v_{e}}{\sqrt {2}}$

Explanation

Lets

$m=$ mass of the particle

$M=$ mass of the earth

$R=$ radius of the earth

$v=$ speed of particle when it reaches the center of the earth

applying conservation of total energy,

$(K.E)_1+(P.E)_1=(K.E)_2+(P.E)_2$

$0+(-\dfrac{GMm}{R})=\dfrac{1}{2}mv^2+(-\dfrac{3GMm}{2R})$

$-\dfrac{GM}{R}+\dfrac{3GM}{2R}=\dfrac{1}{2}v^2$

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

$v=\sqrt{\dfrac{GM}{R}}$ . . . . (1)

We know that the escape velocity of the particle from the earth is given by,

$v_e=\sqrt{\dfrac{2GM}{R}}$ . . . . .(2)

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

$v_e=\sqrt{2} v$ (from equation 1)

$v=\dfrac{v_e}{\sqrt{2}}$

The correct option is D.

1443265_1176877_ans_cf87a3b6ad704a718f4777dd90eba891.png

A tunnel is dug along a diameter of the planet. A particle is dropped from the surface of the planet and reaches the center of the planet with speed

$V$ . The escape velocity from surface of planet is

0%
$\sqrt {2}V$
0%
$2\ V$
0%
$\sqrt {5}V$
0%
$\dfrac {V}{2}$

The magnitude of gravitational potential energy of a body at a distance 'r' from the center of the earth is V. Its weight at a distance '2r' from the center of the earth is

0%
$\frac { V }{ r }$
0%
$\frac { V }{ 4r }$
0%
$\frac { V }{ 2r }$
0%
$\frac { 4V }{ r }$

The weight of the body at earth's surface is W. At a depth halfway to the center of the earth, it will be? (assuming uniform density in the earth)

0%
$W$
0%
$\dfrac{W}{2}$
0%
$\dfrac {W}{ 3 }$
0%
$\dfrac { W} {8 }$

Explanation

${ G }^{ 1 }=g\left( 1-\dfrac { d }{ R } \right) \quad \longrightarrow \left( 1 \right)$

Now

$d$ that is depth is

$\dfrac { R }{ 2 }$ substituting

$d=\dfrac { R }{ 2 }$ in eq

$(1)$

${ G }^{ 1 }=g\left( 1-\dfrac { R }{ 2R } \right) =g\left( 1-\dfrac { 1 }{ 2 } \right) =\dfrac { g }{ 2 }$

So,

${ g }^{ 1 }=\dfrac { g }{ 2 }$

Multiplying both sides by

$M$

${ mg }^{ 1 }=\dfrac { mg }{ 2 }$

Weight is nothing but mass

$\times$ gravity

Weight at half way depth

$=\dfrac { W }{ 2 }$

Ratio of time period of satellites orbiting around earth at a distance of about 9r and 4r from earth surface is : ( where r is the radius of earth)

0%
4:9
0%
9:4
0%
27:8
0%
16:27

The graph represented the relation between acceleration due to gravity(g) and depth (d) from the surface of the earth. The value of g at a depth of

$640\ km$ is

0%
$0.98\ {ms}^{2}$
0%
$7.84\ {ms}^{2}$
0%
$8.82\ {ms}^{2}$
0%
$9\ {ms}^{2}$

A planet moves with respect to us that light of

$475 nm$ is observed at

$475.6 nm$ . The speed of the planet is

0%
$206 kms^{-1}$
0%
$378 kms^{-1}$
0%
$108 kms^{-1}$
0%
$100 kms^{-1}$

If the acceleration due to gravity at a height

$h$ from the surface of the earth is

$96\%$ less than its value on the surface, then

$h$ is (where

$R$ is radius of the earth).

0%
$.5*R$
0%
$2R$
0%
$3R$
0%
$4R$

Explanation

the gravitational acceleration at a height

$h$ above earth's surface is calculated as:

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

where,

$M$ is the mass of the earth

$R$ is the radius of the earth

$G$ is the universal Gravitational constant

at the surface,

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

Rearranging

$g_h$ using the above equations as follows

$\therefore g_h = \dfrac{GM}{R^2}\times\dfrac{R^2}{(R+h)^2} = g \left( \dfrac{R}{R+h} \right)^2$

In the given question,

$g_h$ is

$96\%$ less than its value on the surface

$\implies 100-96 = 4\%$ of its value on surface.

$\therefore g_h = 0.04g$

$\therefore g\left( \dfrac{R}{R+h} \right)^2 = 0.04g$

$\therefore \dfrac{R}{R+h} = \sqrt{0.04} = 0.2 = \dfrac{1}{5}$

$\therefore 5R = R+h$

$\therefore h = 5R-R = 4R$

The gravitational acceleration is

$96\%$ less than its value on the surface of the earth at a height of

$4R$ , where

$R$ is the radius of the earth.

When Johannes Kepler developed his laws (or the movement of planetary bodies), one of the law' stated that the orbits of the planets about the sun are

0%
Circular
0%
Elliptical
0%
Sinusoidal
0%
Straight lines

If the distance between two masses is doubled, the gravitational attraction between them Is doubled

0%
Is doubled
0%
Becomes $4\ times$
0%
Is reduced to half
0%
Is reduced to a quarter

Explanation

According to Newton's law of law of gravitational force of gravitational is inversely proportional to the square of distance between the bodies so distance is doubled force of gravitational will reduce by

$\dfrac { 1 }{ 4 }$ .

Gravitational pull is maximum

0%
Underground
0%
Above the earth
0%
On earth's surface
0%
Underwater

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

0%
$2650$
0%
$2430$
0%
$2250$
0%
$2350$

Explanation

Given,

$R=6400km$

$g'=50g$ %

$g'=0.5g$

New gravity at height

$h$ is given by

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

$0.5g=\dfrac{gR^2}{(R+h)^2}$

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

$\dfrac{R+h}{R}=\sqrt{2}=1.4142$

$1+\dfrac{h}{R}=1.4142$

$\dfrac{h}{R}=0.4142$

$h=0.4142R=0.4142\times 6400$

$h=2650km$

The correct option is A.

The acceleration due to gravity at a depth of 4600 km inside the earth.

0%
2.81 $m/{s^2}$
0%
9.8 $m/{s^2}$
0%
7 $m/{s^2}$
0%
7.35 $m/{s^2}$

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

0%
$\dfrac{W}{2}$
0%
$\dfrac{2W}{3}$
0%
$\dfrac{4W}{9}$
0%
$\dfrac{W}{4}$

Explanation

$\text{At the surface}$

$W=\dfrac{GMm}{R^2}......(I)$

$W'=\dfrac{GMm}{(\dfrac{3R}{2})^2}.......(II)$

$\dfrac{W}{W'}=\dfrac{\dfrac{GMm}{R^2}}{\dfrac{GMm}{9R^2/4}}$

$\Rightarrow \dfrac{W}{W'}=\dfrac{9}{4}$

$\Rightarrow W'=\dfrac{4W}{9}$

1959989_1222786_ans_f38c9320a3e941c1837262a9e638b44d.png

A solid sphere is rotating about an axis passing through its centre with period

$T$ . If its volume shrinks to

$\dfrac {1}{27}$ the earlier volume mass remaining same, the new period will be

0%
$\dfrac {T}{9}$
0%
$3T$
0%
$\dfrac {T}{3}$
0%
$9T$

Assuming the earth to have constant density, point out which of the following curves show variation of acceleration due to gravity from the centre of the point to far away from the surface of the earth.

0%
0%
0%
0%
None of these

Explanation

If distance from centre of earth = r then

$g ∝ r$ if

$r < R$

$g ∝ \dfrac{1}{r^2}$ if

$r > R$ .

1995950_1227740_ans_6183ffad4010455582f6e4212b64dd26.png

Consider a planet moving around a star in an elliptical orbit with period T. The area of the elliptical orbit is proportional to

0%
$T^{4/3}$
0%
$T$
0%
$T^{2/3}$
0%
$T^{1/2}$

The weight of an object in the coal mine, sea level, at the top of the mountain are

$W_{1},W_{2}$ and

$W_{3}$ respectively, then :

0%
$W_{1}< W_{2}> W_{3}$
0%
$W_{1}= W_{2}= W_{3}$
0%
$W_{1}< W_{2}< W_{3}$
0%
$W_{1}> W_{2}> W_{3}$

Explanation

Lets

$m=$ mass of an object

The weight of an object depends on the acceleration due to gravity is given by

$W=mg$

The weight of an object in the coal mine,

$W_1=mg(1-\dfrac{d}{R})$ . . . . .(1)

where,

$R=$ radius of the earth

$d=$ depth of coil mine

The weight of an object at the sea level,

$W_2=mg$ . . . . . . . (2)

The weight of an object at the top of the mountain

$W_3=mg(1-\dfrac{2h}{R})$ . . . .(3)

From equation (1), (2) and (3), we conclude that

$W_1<W_2>W_3$

The correct option is A.

The tilt of Earth is _____ degrees.

0%
$90$
0%
$23.5$
0%
$0$
0%
$35.2$

Spot the wrong statement:
The acceleration due to gravity g decreases if

0%
We go down from the surface of the earth towards its centre
0%
We go up from the surface of the earth
0%
We go from the equator towards the poles on the surface of the earth
0%
The rotational velocity of the earth is increased

If the acceleration due to gravity, g, is 10

$m/s ^{2}$ at the surface of the earth (radius 6400 km), then at a height of 1600 km the value of g will be ( in

$m/s ^{2}$ )

0%
6.4
0%
5
0%
7.5
0%
2.5

Explanation

Let $g_e= 10 m/s$

$R_e =6400m$

$g'=?$

hence by formula

$g'-g_e = \dfrac{1}{2}\times \dfrac{h}{R_e}$

thus g=-4.9 m\s

The change in the value

$'g'$ at a height

$'h'$ above the surface of the earth is the same as at a depth

$'d'$ below the surface of earth. When both

$'d'$ and

$'h'$ are much smaller than the radius of earth, then which one of the fol lowing is correct?

0%
$d=h$
0%
$d=2h$
0%
$d=\dfrac {2h}{2}$
0%
$d=\dfrac {h}{2}$

Explanation

Acceleration due to gravity at a depth 'd' from earth surface is:

$g_d=g\left(1-\dfrac{d}{R}\right)$

Acceleration due to gravity at height 'h' from earth surface is: he is very much smaller than R.

$g_h=g\left(1-\dfrac{2h}{R}\right)$

$g_h=g_d$

By solving it,

$d=2h$ .

The value of acceleration due to gravity on earth surface-

0%
is directly proportional to density of earth
0%
is inversely proportional to density of earth
0%
does not depend on density of earth
0%
none of these

Acceleration due to gravity at surface of a planet is equal to that at surface of the earth and density is

$1.5$ times that of earth. if radius of earth is

$R$ , radius of planet is

0%
$\dfrac {R }{ 1.5 }$
0%
$\dfrac { 2 }{ 3 } R$
0%
$\dfrac {9 }{ 4 } R$
0%
$\dfrac {4 }{ 9 } R$

Explanation

Gravity at surface of a planet

$=$ gravity at surface of the earth.

${ g }_{ p }={ g }_{ e }$

${ P }_{ 1 }={ 1.5P }_{ e }$

Radius of earth

$=R$

$g=\dfrac { G\left( P\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) }{ { R }^{ 2 } }$

It's given,

${ P }_{ P }=1.5{ P }_{ E }$

$\dfrac { G\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 }\times { P }_{ e } }{ { R }^{ 2 } } =\dfrac { G\times \dfrac { 4 }{ 3 } \pi \times { \left( { R }^{ 1 } \right) }^{ 3 }\times { P }_{ P } }{ { \left( { R }^{ 1 } \right) }^{ 2 } }$

or,

$R.{ P }_{ e }={ R }^{ 1 }{ P }_{ P }$

or,

${ R }^{ 1 }=R\times \dfrac { { P }_{ P } }{ { P }_{ P } } =\dfrac { R }{ 1.5 } =\dfrac { 2R }{ 3 }$ .

If the radius of earth shrinks by 1.5 % ( mass remaining same ), then the value of gravitational acceleration changes by

0%
2 %
0%
-2 %
0%
3 %
0%
-3 %

A body weighs 144 N at the surface of earth. When it is taken to a height of

$h=3R$ , where R is radius of earth, it would weigh

0%
48 N
0%
36 N
0%
16 N
0%
9 N

Identify the incorrect statement about a planet revolving around Sun

0%
The gravitational attraction provides the centripetal force for a revolving planet
0%
The total energy of a planet is always negative
0%
The total energy of a planet is always more than potential energy of the system
0%
Kinetic energy of revolving planet is sometimes zero

Explanation

The total energy of a planet is always negative.

The total energy of a planet is always more than potential energy of the system.

kinetic energy of revolving planet is sometimes zero.

But, the gravitational attraction provides the centripetal force for a revolving planet is incorrect.

Hence, Option

$A$ is incorrect statement.

$g_{1}$ is the acceleration due to gravity on the surface of earth,

$g_{2}$ that at a height

$h$ (

$h <<$ radius of earth) and

$g_{3}$ that at a depth

$h$ , then their ascending order is

0%
$g_{3}<g_{2}<g_{1}$
0%
$g_{3}<g_{1}<g_{2}$
0%
$g_{1}<g_{2}<g_{3}$
0%
$g_{2}<g_{3}<g_{1}$

Which of the following graphs shows the variation of acceleration due to gravity

$g$ at height

$h$ from the surface of earth?

Assume (

$h$ <<<<

$Re$ )

0%
$a$
0%
$b$
0%
$c$
0%
$d$

The value of

$'g'$ reduces to half of its value at surface of earth at a height

$'h'$ , then

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

In the figure it is shown that the velocity of lift is $2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$ while string ins winding on the motor shaft with velocity $2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$ and shaft A is moving downward with velocity $2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$ with respect lift, then find out the velocity of block B

0%
$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \uparrow$
0%
$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \downarrow$
0%
$4\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \uparrow$
0%
None of these

Explanation

Lift

$=2$ m/s

velocity

$=2$ m/s

down ward velocity

$=2$ m/s

$\therefore$ velocity of Block

$B=\uparrow 2+2\uparrow -2\downarrow$

$=4-2$

$=2m/s\uparrow$

1210473_1244686_ans_eb65550a16d74efebd7a7b90d50b3801.png

Choose the correct statement:

0%
The dimensional formula for $G$ is ${M}^{-1}{L}^{3}{T}^{-2}$
0%
$G$ is independent of medium
0%
$F=G\cfrac{{m}_{1}{m}_{2}}{{r}^{2}}$
0%
All of the above

The value of universal gravitational constant on earth for a particle of mass 5 kgs is

0%
$6.67\times { 10 }^{ -11 }$
0%
$6.67\times { 10 }^{ -7 }$
0%
$5\times 6.67\times { 10 }^{ -11 }$
0%
$6.67\times { 10 }^{ -23 }$

Explanation

A universal gravitational constant is a constant number that does not depends on the masses of any planets or objects. And it is equal to

$G=6.67\times 10^{-11}$

$Nm^2kg^{-2}$

Option A

The value of acceleration due to gravity on Mount Everest is

0%
$g$
0%
$>g$
0%
$<g$
0%
$zero$

The value of universal gravitational constant depend upon:

0%
Nature of material of two bodies
0%
Heat constant of two bodies
0%
Acceleration of two bodies
0%
None of these

Explanation

Every object in the universe attracts every another object 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.

$F\propto \dfrac{m_1m_2}{r^2}$

$F=G \dfrac{m_1m_2}{r^2}$

where

$G$ is the constant of proportionality and is called the universal gravitation constant.

$G$ does not depend on anything.

If the earth were to suddenly contract to half its present size, without any change in its mass, the duration of the new day be

0%
$18\ hours$
0%
$30\ hours$
0%
$6\ hours$
0%
$112\ hours$

Explanation

Given,

$r_2=\dfrac{1}{2}r_1$

$m_1=m_2=m(say)$

$T_1=24hr$

By the conservation of angular momentum,

$I_1\omega _1=I_2\omega _2$ . . . . . . . . .(1)

Where,

$\omega =\dfrac{2\pi}{T}$ angular velocity of the earth

$I=\dfrac{2}{5}mr^2$ moment of inertia of earth.

From equation (1),

$\dfrac{2}{5}mr_1^2.\dfrac{2\pi}{T_1}=\dfrac{2}{5}mr_2^2.\dfrac{2\pi}{T_2}$

$r_1^2.\dfrac{1}{T_1}=r_2^2.\dfrac{1}{T_2}$

$\dfrac{T_2}{T_1}=(\dfrac{r_2}{r_1})^2=(\dfrac{r_1/2}{r_1})^2=\dfrac{1}{4}$

$T_2=T_1.\dfrac{1}{4}$

$T_2=\dfrac{24}{4}hr=6hr$

The correct option is C.

The acceleration due to gravity on a planet is

$1.96\,m/{s^2}$ . If it is safe jump from a height of

$3$ m on the earth, the corresponding height on the planet will be

0%
3 m
0%
6 m
0%
9 m
0%
1.5 m

Find the false statement

0%
Gravitational force acts along the line joining the two interacting particles
0%
Gravitational force is independent of medium
0%
Gravitational force forms an action-reaction pair
0%
Gravitational force does not obey the principle of superposition

A body of mass

$m$ is raised to a height

$h$ above the surface of the earth of mass

$M$ and radius

$R$ until its gravitational potential energy increases by

$\dfrac{1}{3} mgR$ . The value of

$h$ is

0%
$\dfrac{R}{3}$
0%
$\dfrac{R}{2}$
0%
$\dfrac{mR}{(M + m)}$
0%
$\dfrac{mR}{M}$

If the radius of the earth is suddenly contracts to half of its present value, then duration of day will be of

0%
6 hours
0%
12 hours
0%
18 hours
0%
24 hours

The motion of planets in the solar system is an example of the conservation of

0%
Mass
0%
Linear momentum
0%
Angular momentum
0%
Energy

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

0%
2630
0%
2640
0%
2650
0%
2660

A man weighs

$'W'$ on the surface of the earth and his weight at a height

$'R'$ from surface of the earth is ( R is Radius of the earth)

0%
$\cfrac{W}{2}$
0%
$\cfrac{W}{4}$
0%
$W$
0%
$4W$

If retardation produced by air resistance of air resistance is one - tenth of acceleration due to gravity. the time to reach maximum height.

0%
Decrease by 11 percent
0%
Increase by 11 percent
0%
Decrease by 9 percent
0%
Increase by 9 percent

An express train is moving with a velocity

$\mathbf { V } _ { 1 }$ Its driver finds another train is moving on the same track in the same direction with velocity

$v _ { 2 }$ s. To escape collision. driver applies retardation a on the train. The minimum time of escaping collision will be

0%
$t = \frac { v _ { 1 } - v _ { 2 } } { a }$
0%
$t = \frac { v _ { 1 } ^ { 2 } - v _ { 2 } ^ { 2 } } { 2 }$
0%
None
0%
Both

Explanation

As the trains moving in the same direction so the initial relative speed

$(V_1-V_2)$ and by applying retardation final relative speed become zero.

From

$\begin{array}{l} V=u-at \\ O=\left( { { V_{ 1 } }-{ V_{ 2 } } } \right) -at \\ t=\dfrac { { { V_{ 1 } }-{ V_{ 2 } } } }{ a } \end{array}$

A battery of weight

$60kg-wt$ is bought from earth surface to a space station at height equal to

$R_e$ , the weight of battery at their.

0%
$45kg-wt$
0%
$60kg-wt$
0%
$30kg-wt$
0%
$15kg-wt$

Ratio of gravitational force acting on body at height

$1600\ km$ to gravitational force acting on the same body on Earth's surface

$(R = 6400\ km)$ .

0%
$5 : 4$
0%
$15 : 16$
0%
$4 : 5$
0%
$16 : 25$

Escape velocity is given by :

0%
$\sqrt{2 gR}$
0%
$\sqrt {2g/R}$
0%
$\sqrt{2} gR$
0%
$\sqrt[2]{gR}$

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

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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