On the surface of earth acceleration due to gravity is g and gravitational potential is V. Column-I Column-II a) At height h = R, value of g (p) decreases by a factor 1/4 (b) At depth h = R/2 value of g (q) decreases by a factor 1/2 (c) At height h = R, of V (r) Increases by a factor 11/8 (d) At depth h = R/2 of V (s) Increases by a factor 2 (t) None

(a - p), (b - q), (c - s), (d - t)

(a - q), (b - p), (c - t), (d - s)

(a - t), (b - s), (c - p), (d - q)

Decreases with increase in depth

Increases with increase in depth

Remains same regardless of the depth

Depends upon the mass of the object

MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 6

A body of mass

$m$ is lifted up from the surface of earth to a height three times the radius of the earth

$R$ .The change in potential energy of the body is

0%
$3mgR$
0%
$\displaystyle\frac{5}{4}mgR$
0%
$\displaystyle\frac{3}{4}mgR$
0%
$2mgR$

Explanation

Hint: Use formula of change in potential energy

$Step - 1: Put\ formula\ of\ change\ in\ potential\ energy\ with\ height$

Change in potential energy is given by the formula:

$∆U = \dfrac{m\ g\ h}{1\ +\ \dfrac{h}{R}}$

Now, if body is lifted to height three times to the radius of earth, then h = 3 R.

$Step - 2: Put\ values\ in\ formula\$

On putting value in formula,

$∆U = \dfrac{m\ g\ (3\ R)}{1\ +\ \dfrac{3\ R}{R}} = \dfrac{3}{4} m g R$

$Answer:$

Hence, option C is the correct answer.

At some planet '

$g$ ' is

$1.96 m/s^2$ . If it is safe to jump from a height of

$2m$ on earth, then what should be corresponding safe height for jumping on that planet

0%
$5m$
0%
$2m$
0%
$10m$
0%
$20m$

Explanation

At earth,

$g_e=9.8 m/s^2 \quad h_e=2 m$

At other planet,

$g_p=1.96 \quad h=?$

$\implies \dfrac{g_p}{g_e}= \dfrac{1}{5}$

Now at some planet g reduces by

$5$ times, thus safe height will increases by

$5$ times.

Hence

$h= 5\times 2= 10 m$

On the surface of earth acceleration due to gravity is g and gravitational potential is V.

Column-I	Column-II
a) At height h = R, value of g	(p) decreases by a factor 1/4
(b) At depth h = R/2 value of g	(q) decreases by a factor 1/2
(c) At height h = R, of V	(r) Increases by a factor 11/8
(d) At depth h = R/2 of V	(s) Increases by a factor 2 (t) None

0%
(a - p), (b - q), (c - s), (d - t)
0%
(a - q), (b - p), (c - t), (d - s)
0%
(a - t), (b - s), (c - p), (d - q)
0%
None of these

Explanation

$\text{Value of g at a distance 'r' from center of earth}=\dfrac{GM}{r^2}$

$\text{'g' at surface}=\dfrac{GM}{R^2}$

$'g' \text{at height}\ 'R' \text{from surface}=\dfrac{GM}{(2R)^2}$

$\text{Hence decreased by a factor of}\ \dfrac{1}{4}.$

$\text{Dependence of 'g' on the depth h is}\ g_h=\dfrac{4\pi}{3}G\rho h$

$\text{Hence decreases by a factor of}\ \dfrac{1}{2}\ \text{at}\ h=\dfrac{R}{2}.$

$\text{Gravitational potential varies with distance 'r' from center of earth as}-\dfrac{GM}{r}$

$\text{Hence at surface it has value}-\dfrac{GM}{R}$

$\text{At a height 'R' it has value}-\dfrac{GM}{2R}$

$\text{Hence it has 'increased' by a factor 2, since potential is negative.}$

If velocity of a satellite is half of escape velocity, then distance of the satellite from earth surface will be.

0%
$6400 km$
0%
$12800 km$
0%
$6400 \sqrt 2 km$
0%
$\dfrac {6400}{\sqrt 2}km$

Explanation

Orbital velocity ,

$v=\sqrt{\dfrac{GM}{r}}$ where

$r$ is the radius of orbit

For

$v=\dfrac{v_e}{2} = \dfrac{1}{2}\sqrt{\dfrac{2GM}{R}}$

$\dfrac{1}{2}\sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{GM}{r}} \implies r=2R$

$r= h+R$ , thus

$h= R= 6400 km$

Mass of a planet is

$5 \times 10^{24}$ kg and radius is

$6.1 \times 10^6$ m. The energy needed to send a

$2$ kg body into space from its surface, would be.

0%
$9$ joule
0%
$18$ joule
0%
$2.2 \times 10^8$ joule
0%
$1.1 \times 10^8$ joule

Explanation

Energy required to send a body into the space from the surface of planet is equal to the negative of potential energy at surface.

At surface

$P.E = -\dfrac{GMm}{R}$

Thus energy required

$E= \dfrac{GMm}{R}$

Given:

$M= 5 \times 10^{24} kg m= 2 kg R= 6.1 \times 10^6 m G= 6.67 \times 10^{-11} Nm^2/kg^2$

$\implies E=1.1 \times 10^{8} Joule$

Gravitational potential difference between surface of a planet and a point situated at a height of

$20m$ above its surface is

$2 joule/kg$ . If gravitational field is uniform, then the work done in taking a

$5kg$ body of height

$4$ meter above surface will be :-

0%
$2J$
0%
$20J$
0%
$40J$
0%
$10J$

Explanation

The potential difference for a distance of

$20 m$ is

$2 \ joule/kg$ .

Now work done

$=m \times$ potential difference.

Thus work done for a distance of

$4 m$ for a mass

$5 kg$ ,

$W= \dfrac{4}{20} \times 2 \times 5$

$\implies W= 2 J$

Potential energy of a

$3 kg$ body at the surface of a planet is

$54J$ then escape velocity will be.

0%
$18 m/s$
0%
$162 m/s$
0%
$36 m/s$
0%
$6 m/s$

Explanation

Magnitude of gravitational potential energy at surface,

$|P.E|= \dfrac{GMm}{R}$

For

$m= 3 kg$ gives

$54= \dfrac{GM(3)}{R}$

$\implies \dfrac{GM}{R} = \dfrac{54}{3}$

Now escape velocity,

$v_e= \sqrt{\dfrac{2GM}{R}}= \sqrt{\dfrac{2\times 54}{3}}$

$\implies v_e= 6 m/s$

An object weighs

$10 N$ at the north pole of theearth. In a geostationary satellite distance

$7R$ from the centre of the earth (of radius

$R$ ), the true weight and the apparent weight are.

0%
$0 N, 0 N$
0%
$0.2 N, 0 N$
0%
$0.2 N, 9.8 N$
0%
$0.2 N, 0.2 N$

Explanation

Acceleration due to gravity at a height h from the surface ,

$g'= g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$

For radius of orbit

$= 7R \implies h=6R$

$mg'= mg\bigg[1+\dfrac{6R}{R}\bigg]^{-2}$

$mg'= 10 \times [7]^{-2} N$

$\implies W'= .2 N$

Also apparent weight of an object will bo zero.

A body of mass

$m$ is situated at distance

$4R_e$ above the earth's surface, where

$R_e$ is the radius of earth how much minimum energy be given to the body so that it may escape

0%
$mgR_e$
0%
$2mgR_e$
0%
$\dfrac {mgR_e} {5}$
0%
$\dfrac {mgR_e} {16}$

Explanation

Potential energy of the body at a distance 4

$R_e$ from the surface of earth

$U=-\dfrac{mgR_e}{1+n/R_e}$

$=-\dfrac{mgR_e}{1+4}=-\dfrac{mgR_e}{5}$ As

$h=4R_e$

So, minimum energy required to escape the body will be

$\dfrac{mgR_e}{5}$

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

Assertion : Radius of circular orbit of a satellite is made two times, then it areal velocity will also become two times.

Reason : Areal velocity is given as

$\dfrac {dA}{dt}=\dfrac {L}{2m}=\dfrac {mvr}{2m}$

0%
If both assertion and reason are true and the reason is the correct explanation of the assertion
0%
If both assertion and reason are true but reason is not the correct explanation of the assertion
0%
If assertion is true but reason is false
0%
If assertion is false but reason is true

Explanation

Areal velocity

$\dfrac{dA}{dt}= \dfrac{L}{2m}=\dfrac{mvr}{2m}$

As there is no external torque, thus

$L=constant$

Now if radius of orbit increases by 2 times then the velocity of satellite will get reduced by 2 times so as to make

$L$ to be constant.

Hence overall, increasing

$r$ has no effect on areal velocity.

So the assertion is false.

A space shuttle is launched in a circular orbit near the earth's surface. The additional velocity be given to the space - shuttle to get free from the influence of gravitational force, will be.

0%
$1.52\ km/s$
0%
$2.75\ km/s$
0%
$3.28\ km/s$
0%
$5.18\ km/s$

Explanation

Orbital velocity of satellite revolving near earth's surface,

$v= \sqrt{\dfrac{GM}{R}}$

$v= \sqrt{gR}$ (using

$GM=gR^2$ )

$v= \sqrt{9.8\times 6400\times 1000}$

$\implies v= 7.919 km/s$

Escape velocity

$v_e= \sqrt{2gR}= \sqrt{2(9.8)(6400\times 1000)}= 11.2 km/s$

Thus additional velocity required

$= 11.2-7.919 km/s= 3.28 km/s$

If the change in the value of '

$g$ ' at a height

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

$d$ below it. When both

$d$ and

$h$ are much smaller than the radius of earth, then

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

Explanation

the value of 'g' at a height

$h$ above the surface of the earth,

$g_h=g(1-\dfrac{2h}{R})$
the value of 'g' at a depth d,

$g_d=g(1-\dfrac{d}{R})$
But,

$g_h=g_d$

$\therefore$

$1-\dfrac{2h}{R}=1-\cfrac{d}{R}$
or,

$d=2h$

The escape velocity for a planet is

$\displaystyle v_{e}$ A particle starts from rest at a large distance from the planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre Its speed at the centre of the planet will be

0%
$\displaystyle \sqrt{1.5}v_{e}$
0%
$\displaystyle \frac{v_{e}}{\sqrt{2}}$
0%
$\displaystyle v_{e}$
0%
zero

Explanation

From mechanical energy conservation 0 + 0 = $\displaystyle \frac{1}{2}mv^{2}-\frac{3GMm}{2R}\: \: \Rightarrow \: \: v=\sqrt{\frac{3GM}{R}}=\sqrt{1.5}v_{e}$

The ratio of the radii of the planets

${P}_{1}$ and

${P}_{2}$ is

$k$ . The ratio of gravitational field intensity at their surface is

$r$ then the ratio of the escape velocities from them will be -

0%
$kr$
0%
$\sqrt{kr}$
0%
$\displaystyle\sqrt{\frac{k}{r}}$
0%
$\displaystyle\sqrt{\frac{r}{k}}$

Explanation

We are given that:

$\dfrac{\dfrac { G{ M }_{ 1 } }{ { R }_{ 1 }^{ 2 } }} {\dfrac { G{ M }_{ 2 } }{ { R }_{ 2 }^{ 2 } } }=r$

$\\ \dfrac { { R }_{ 1 } }{ { R }_{ 2 } } =k$ .

And the escape velocity is given by:

$v=\sqrt { \dfrac { 2GM }{ R } }$ where

$M$ is the mass of the planet and

$R$ is its radius.

So,

$\dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\sqrt { \dfrac { { M }_{ 1 } }{ { M }_{ 2 } } /\dfrac { { R }_{ 1 } }{ { R }_{ 2 } } } =\sqrt { \dfrac { r{ k }^{ 2 } }{ k } } =\sqrt { rk }$ .

The dependence of acceleration due to gravity

$g$ on the distance

$r$ from the centre of the earth, assumed to be a sphere of radius

$R$ of uniform density is as shown in figures below. The correct figure is

Explanation

The acceleration due to gravity at a depth d below surface of earth is

$\displaystyle g' = \frac{GM}{R^2} \left( 1 - \frac{d}{R} \right) = g \left( 1 - \frac{d}{R} \right)$

$g' = 0$ at

$d = R$
i.e., acceleration due to gravity is zero at the centre of earth.
Thus, the variation in value

$g$ with

$r$ is for,

$r > R$ ,

$\displaystyle g' = \frac{g}{ \left( \displaystyle 1 + \frac{h}{R} \right)^2} = \frac{gR^2}{r^2} \Rightarrow g' \propto \frac{1}{r^2}$
Here,

$R + h = r$
For

$\displaystyle r < R, g' = g \left( 1 - \frac{d}{R} \right) = \frac{gr}{R}$
Here,

$R - d = r$

$\Rightarrow g' \propto r$
Therefore, the variation of

$g$ with distance from centre of the earth will be as shown in the figure.

$A$ is the areal velocity of planet of mass

$M$ . its angular momentum is

0%
$M$
0%
$2MA$
0%
$A^2M$
0%
$AM^2$

Explanation

$\textbf{Hint: First calculate $\omega$ and Moment of inertia of planet is, $I=M{R}^{2}$.}$

$\textbf{Step 1: Calculating the angular velocity.}$

Given that the planet has an aerial velocity of A, Aerial velocity is given by,

$A=\dfrac{Area swept by radius vector}{Time taken}$

$\Rightarrow A=\dfrac{\pi {{R}^{2}}}{T}$ (Where R is the radius of the planet)

We know that, $T=\dfrac{2\pi }{\omega }$

$\Rightarrow A=\dfrac{\omega {{R}^{2}}}{2}$

$\Rightarrow \omega =\dfrac{2A}{{{R}^{2}}}$

$\textbf{Step 2: Calculating the angular momentum.}$

Angular momentum of the planet will be, $L=I\omega =M{{R}^{2}}\times \dfrac{2A}{{{R}^{2}}}$

$\Rightarrow L=2MA$

$\textbf{Thus, option (B) is correct.}$

The velocity of a planet revolving around the sun at three different times of a year is shown in the figure. Which among the following alternatives is correct ?

0%
$v_2\, =\, \displaystyle \frac{v_1\, +\, v_3}{2}$
0%
$v_1\, =\, \displaystyle \frac{v_2\, +\, v_3}{2}$
0%
$v_1\, >\, v_2\, >\, v_3$
0%
$v_1\, <\, v_2\, <\, v_3$

Explanation

Kepler's law states that angular momentum is conserved.

$L = constant$

$v_{1} r_{1} = v_{2}r_{2} = v_{3}r_{3}$
r is the position vector from one point to the sun.
As we can see from the ellipse:

$r_{3} > r_{2} > r_{1}$
Hence,

$v_{1} > v_{2} > v_{3}$
Option C is correct.

The variation of

$g$ with height or depth (

$r$ ) is shown correctly by the graph in figure (where

$R$ is radius of the earth).

Explanation

For depth:

$g \propto (R-d)$
For height:

$g \propto \cfrac{1}{R^2}$
Hence, the graph will be linear for depth from centre of earth

$(R=0)$ till the surface of earth

$(R)$ and then parabolic with increase in height from earth.

Above the earth's surface, the variation of g w.r.t. the height (r) is correctly represented by which of the given proportionalities

0%
$g \propto \dfrac {1}{r^2}$
0%
$g \propto r$
0%
$g \propto \ r^2$
0%
$g \propto r^o$

Explanation

The variation of g with the height(r)

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

$\Rightarrow g_r$

$\alpha$

$\dfrac{1}{r^2}$

Hence correct answer is option

$A$

The SI unit of G is.

0%
$N^2m^2/kg$
0%
$Nm^2/kg$
0%
$N \ ml \ kg$
0%
$Nm^2/kg^2$

Explanation

$\textbf{Correct option: Option (D).}$

$\textbf{Explanation:}$

$\bullet$ Gravitational force is the universal force of attraction experienced by all matters.

$\bullet$ It is given by, $F=\dfrac{GMm}{{{R}^{2}}}$ (Where G is the universal gravitational constant whose value remains constant throughout the space, and M and m are the masses of the two objects and R be the distance between the bodies.)

Then, $G=\dfrac{F{{R}^{2}}}{Mm}$

For a force of 1 Newton between bodies of mass 1 kg and 1 meter apart, G will be

$\bullet$ S.I unit of G

$\Rightarrow G=\dfrac{N\times {{m}^{2}}}{kg\times kg}$

$\Rightarrow G=N{{m}^{2}}/k{{g}^{2}}$

$\textbf{Thus, option (D) is correct.}$

Where will 'g' be greatest when one goes from the centre of earth to an altitude equal to the radius of the earth?

0%
at the surface of earth
0%
at the centre of earth
0%
at the highest point
0%
none of the above

Explanation

Let the radius of earth be

$R$ and density

$\rho$

The value of

$g$

for

$0<r<R$ =

$\frac{G\rho r}{3}$ (directly proportional to

$r$ )

$r>R$ =

$\frac{G\rho R^{3}}{3r^{2}}$ (inversely proportional to r)

maximum value of

$g$ is at

$r=R$

$g=\frac{G \rho R}{3}$

The weight of an object would be minimum when it is placed :

0%
At the North Pole
0%
At the South Pole
0%
At the Equator
0%
At the Centre of the Earth

Explanation

$Answer:-$ D

Value of g at any point on surface of earth is given by:-

${ g }^{ \prime }=g-{ \omega }^{ 2 }R\cos { \varphi }$

At equator

$cos\varphi =1\ so\ { g }^{ \prime }=g-{ \omega }^{ 2 }R$

Value of

$\cos { \varphi } <1$ at every point other than equator.

So,

${ g }^{ \prime }$ will be more at any point other than equator relative to equator.

Variation of

$g$ with depth is given by:

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

So, when depth d is maximum g is minimum. So, weight (mg) will also be minimum.

In the motion of the planets,

0%
The angular velocity is constant
0%
The linear momentum is constant
0%
The angular momentum is constant
0%
None of the above

If a body is sent with a velocity of ............... km

$\displaystyle { sec }^{ -1 }$ , it would leave the earth forever.

0%
$11.9$
0%
$11.6$
0%
$11.4$
0%
$11.2$

Explanation

$Answer:-$ D

Escape velocity is the minimum speed needed for an object to escape from the gravitational attraction of a massive body. The escape velocity from Earth is about 11.186 km/s (40,270 km/h; 25,020 mph) at the surface. More generally, escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero.

And it can be calculated by :-

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

From the centre of the earth to the surface of the earth, the relation between the value of g and distance (r) represented as a proportionality, is given by

0%
$g \propto \dfrac {1}{r^2}$
0%
$g \propto r$
0%
$g \propto r^2$
0%
$g \propto r^o$

Explanation

The variation of g with the depth(r)

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

$\Rightarrow g_r$

$\alpha$

$r$

Hence correct answer is option

$B$

The maximum weight of a body on earth is

0%
at the centre of the earth
0%
inside the earth
0%
on the surface of the earth
0%
above the surface of the earth

The value of

$g$

0%
Increases with increase in depth
0%
Decreases with increase in depth
0%
Remains same regardless of the depth
0%
Depends upon the mass of the object

Explanation

$Answer:-$ B

Variation of g with depth is given by:

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

where,

$g$ is acceleration due to gravity on the surface,

$d$ is depth,

$R$ is radius of earth

so we can see that as value of d increases

$g_d$ decreases.

An object is dropped at the surface of the earth from the height of

$3600$ km. Calculate the ratio of the weight of the body at that height and on the surface of the earth.

0%
$1.34$
0%
$2.44$
0%
$6.25$
0%
$12.32$

Explanation

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

$\therefore \dfrac{mg_{surface}}{mg_{height}}=\left(\dfrac{R+H}{R} \right)=\left(\dfrac{10000}{6400}\right)$

$\implies 2.44$

The variation of g with height or depth (r) is shown correctly by the graph in the figure (where R

$=$ radius of the earth ),

Explanation

for

$r\ge R;\quad g=\cfrac { GM }{ { r }^{ 2 } }$ - eq. 1

For r<R; g=

$\cfrac { GMr }{ { R }^{ 3 } }$ - eq.2

According to relations 1 and 2

the g vs r curve will be like

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

Explanation

Unit of

$G$ is

$Nm^2kg^{-2}$ as

$1N=kgms^{-2}$ .

$G$ can be written as

$kgms^{-2}\times m^2kg^{-2}=m^3kg^{-1}s^{-2}$

Where will g be greatest when one goes from the centre of earth to an altitude equal to radius of earth?

0%
at the surface of earth
0%
at the centre of earth
0%
at the highest point
0%
none of the above

Explanation

Let at any height

$h$ gravity due to earth will be

$g$ .

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

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

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

$\therefore$ As

$h$ increases

$g$ decreases.

$\Rightarrow$ At

$h=0$ ,

$g$ will be maximum,

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

So ,

$g$ will be maximum at the surface of the earth.

Maximum weight of the body is

0%
at the centre of the eath
0%
inside the earth
0%
on the surface of the earth
0%
above the the surface of the earth

Explanation

$\textbf{Hint: Formula for gravitational acceleration is g}$ = $\dfrac{GM}{{{R}^{2}}}$

$\textbf{Step 1: Weight of the body}$

Weight of a body is (W)=mg.

Since, mass (m) is constant, ‘g’ is the variable force, also called gravitational force.

$\textbf{Step 2: Utilising the formula for gravitational acceleration}$

The effect on gravitational force can be found by considering the formula

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

where;

G = Gravitational constant ( $6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$ )

M = Mass of Earth

R = distance between centre of the earth and the body.

$\textbf{Step 3: Finding the effect of ‘g’ above the surface}$

Now, when the body goes away from the centre of the Earth, ‘R’ increases.

Since $g\propto \dfrac{1}{R}$ , the gravitational force decreases.

Hence, gravitational force above the surface of the Earth is lesser than at the surface.

$\textbf{Step 4: Finding the effect of ‘g’ below the surface}$

When the body goes towards the centre of the Earth, although ‘R’ decreases; ‘M’ decreases as the mass of earth below the body is lesser/decreasing.

Since $g\propto M$ , the gravitational force decreases.

Hence, gravitational force below the surface of the Earth is lesser than at the surface.

$\textbf{Step 5: Finding the effect of ‘g’ at the centre}$

When the body is at the centre of the Earth, M=0.

⸫ g = 0

Hence, gravitational force at centre of Earth is 0.

$\textbf{Step 6: Conclusion}$

Since ‘g’ is maximum for body at the surface of the earth, then maximum weight of the body (W) will also be at the surface of the Earth.

$\textbf{Answer:}$

$\textbf{Hence, the correct option is (C) on the surface of the Earth}$

If R is the radius of the earth and g is the acceleration due to gravity on the earth's surface, then mean density of the earth is

0%
$\dfrac { 4\pi G }{ 3gR }$
0%
$\dfrac { 3\pi R }{ 4gG }$
0%
$\dfrac { 3 G}{ 4\pi RG }$
0%
$\dfrac { pi g }{ 12RG }$

What will be acceleration due to gravity on the surface of moon if its radius is

$\frac { 1 }{ 4 }$ th the radius of the earth and its mass is

$\frac { 1 }{ 80 }$ th the mass of the earth?

0%
$\frac { g }{ 2 }$
0%
$\frac { g }{ 3 }$
0%
$\frac { g }{ 7 }$
0%
$\frac { g }{ 5 }$

Explanation

We know,

$g=\frac { { GM }_{ e } }{ { R }_{ e }2 } \\ and\quad g\quad '=\frac { { GM }_{ m } }{ { R }_{ m }2 } \\ \frac { g\quad ' }{ g } =\frac { M_{ m } }{ M_{ e } } \left( \frac { { R }_{ e } }{ { R }_{ m } } \right) ^{ 2 }\\ =\frac { 1 }{ 80 } \times \left( \frac { 4 }{ 1 } \right) ^{ 2 }=\frac { 1 }{ 5 } \\ \Rightarrow g' = \frac { g }{ 5 }$

SI unit of G is.

0%
$N^2 -m^2/kg$
0%
$N -m^2/kg$
0%
$N -m/kg$
0%
$N -m^2/kg^2$

Explanation

Hint: using the formula of gravitional constant

Step1:We know that

$\mathrm{F}=\mathrm{G} \times \dfrac{M_{1} M_{2}}{r^{2}} \\$

$\mathrm{~F} \times \mathrm{r}^{2}=\mathrm{G} \times \mathrm{M}_{1} \mathrm{M}_{2} \\$

$\mathrm{~F} \times \dfrac{r^{2}}{M_{1} M_{2}}=\mathrm{G} \\$

$\mathrm{G}=\mathrm{F} \times \dfrac{r^{2}}{M_{1} M_{2}}$

SI Unit of

$\text { Force }=\mathrm{F}=\text { Newton } \\$

$\text { Distance }=\mathrm{R}=\mathrm{m} \\$

$\text { Mass }=\mathrm{M}=\mathrm{kg}$

Step2: Now,

$\mathrm{G}=\mathrm{F} \times \dfrac{r^{2}}{M_{1} M_{2}}$

$\mathrm{G}=\dfrac{\text { Newton } \times \text { Meter }^{2}}{\text { Kilogram } \times \text { Kilogram }}$

$\mathrm{G}=\dfrac{\text { Newton Meter }^{2}}{\text { Kilogram }^{2}}$

$\mathrm{G}=\mathrm{Nm}^{2} / \mathrm{kg}^{2}$

So, SI Unit of $\mathrm{G}$ is $\mathrm{Nm}^{2} / \mathrm{kg}^{2}$

$\textbf{Thus, option (D) is correct.}$

At what height, is the value of

$g$ half that on the surface of earth? (

$R =$ radius of the earth)

0%
$0.414\ R$
0%
$R$
0%
$2\ R$
0%
$3.5\ R$

Explanation

$g\quad '=\frac { { gR }^{ 2 } }{ { r }^{ 2 } } =\frac { g }{ 2 } \\ \Rightarrow \quad { r }^{ 2 }={ 2R }^{ 2 }\\ \Rightarrow r=R\sqrt { 2 } =R+h\\ \Rightarrow h=R(\sqrt { 2 } -1)\\ =R(1.414-1)=0.414R$

The value of g near the earth's surface is.

0%
$8.9\ ms^{-2}$
0%
$8.9\ ms^{-1}$
0%
$9.8\ ms^{-2}$
0%
$9.8\ ms^{-1}$

Explanation

Its value is

$9.8\ ms^{-2}$ on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is

$9.8\ ms^{-2}$ . When discussing the acceleration of gravity, it was mentioned that the value of

$g$ is dependent upon location.

A body of mass

$m$ is raised to a height

$10R$ from the surface of the earth, where

$R$ is the radius of the earth. The increase in potential energy is (

$G =$ universal constant of gravitation,

$M =$ mass of the earth and

$g =$ acceleration due to gravity)

0%
$\dfrac{GMm}{11R}$
0%
$\dfrac{GMm}{10R}$
0%
$\dfrac{mgR}{11G}$
0%
$\dfrac{10GMm}{11R}$

Explanation

PE at the earth surface

$= \dfrac{-GMm}{R}$
PE at a height

$h$ above the earth's surface

$= \dfrac{-GMm}{\left(R+h\right)}$

$\therefore$ Change in PE

$= \dfrac{-GMm}{\left(R+h\right)} - \left(-\dfrac{GMm}{R}\right) = \dfrac{GMmh}{R\left(R+h\right)}$
Substituting

$h = 10R$

$\triangle PE = \dfrac{GMm \times 10R}{R\left(R+10R\right)} = \dfrac{10GMm}{11R}$

Let

$g_h$ and

$g_d$ be the acceleration due to gravity at height h above the earth's surface and at depth d below the earth's surface respectively. If

$g_h=g_d$ , then the relation between h and d is

0%
$d = h$
0%
$d = \dfrac{h}{2}$
0%
$d = \dfrac{h}{4}$
0%
$d = 2h$

Explanation

Let the radius of the earth be

$R$ and acceleration due to gravity at earth surface be

$g$ .

Acceleration due to gravity at height

$h$ above the earth surface,

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

Acceleration due to gravity at depth

$d$ below the earth surface,

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

Now for

$g_h = g_d$ we get,

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

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

$\Rightarrow$

$1 - \dfrac{2h}{R} = 1 - \dfrac{d}{R}$

$\Rightarrow$

$d = 2h$

The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on

0%
Mass of the earth
0%
Mass of the projectile
0%
Radius of the projectile's orbit
0%
Gravitational constant

Explanation

Using this we calculate escape velocity:

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

In this expression we can see mass of earth(M),Radius of orbit(R), Gravitational constant(G). Mass of projectile does not matter.

The magnitude of acceleration due to gravity at an altitude 'h' from the earth is equal to its magnitude at a depth 'd'. Find the relation between 'h'and 'd'. If the 'h'and 'd' both increases by

$50$ %, are the magnitudes of acceleration due to gravity at the new altitude and the new depth equal.

0%
$d \ =\ h$
0%
$d\ =\ 2h$
0%
$\displaystyle h = \frac { d }{ 3 }$
0%
$d\ =\ 4h$

Explanation

${ g }_{ n }=g\left( 1-\cfrac { 2h }{ R } \right) \\ { g }_{ d }=g\left( 1-\cfrac { d }{ R } \right)$

${ g }_{ n }={ g }_{ d }\\ g\left( 1-\cfrac { 2h }{ R } \right) =g\left( 1-\cfrac { d }{ R } \right) \\ \Rightarrow \cfrac { 2h }{ R } =\cfrac { d }{ R } \\ \therefore 2h=d$

A body of mass '

$m$ ' is raised to a height '

$10R$ ' from the surface of the Earth, where '

$R$ ' is the radius of the Earth. The increase in potential energy is ____ . (

$G$ = universal constant of gravitation,

$M$ = mass of earth and

$g$ = acceleration due to gravity).

0%
$\displaystyle\frac{GMm}{11R}$
0%
$\displaystyle\frac{GMm}{10R}$
0%
$\displaystyle\frac{mgR}{11G}$
0%
$\displaystyle\frac{10GMm}{11R}$

Explanation

By conservation of energy, work done = increase in potential energy.

Potential energy at a height

$x$ above the surface of the Earth is

$-\dfrac{GMm}{R+x}$ .

So, increase in potential energy = work done =

$\dfrac { GMm }{ R } -\dfrac { GMm }{ 11R } =\dfrac { 10GMm }{ 11R}$

The height at which the acceleration due to gravity is

$25\%$ of that of the surface of earth is ____________. (R=Radius of the earth)

0%
$h=3R$
0%
$h=2R$
0%
$h=R$
0%
$h=\displaystyle\frac{R}{2}$

Explanation

Acceleration due to gravity at height

$h$ is

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

Let the height be

$h$ where

$g' =25$ % of

$g = \dfrac{g}{4}$

$\Rightarrow$

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

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

$\Rightarrow h = R$

The change in the gravitational potential energy when a body of mass m is raised to a height

$nR$ above the surface of the Earth is (Here R is the radius of the earth)

0%
$\begin{pmatrix}\dfrac{n}{n+1}\end{pmatrix}mgR$
0%
$\begin{pmatrix}\dfrac{n}{n-1}\end{pmatrix}mgR$
0%
$nmgR$
0%
$\dfrac{mgR}{n}$

Explanation

The acceleration due to gravity varies with height as

$g'=\dfrac{g}{\begin{pmatrix}1+\dfrac{h}{R}\end{pmatrix}}$

Hence,

$h=nR\Rightarrow g'=\dfrac{g}{\begin{pmatrix}1+\dfrac{nR}{R}\end{pmatrix}}=\dfrac{g}{(1+n)}$

The change in potential energy is equal to

$\Delta U=mgh=\dfrac{mg(nR)}{(1+n)}=\begin{pmatrix}\dfrac{n}{n+1}\end{pmatrix}mgR$

$g$ is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass

$m$ raised from the earth's surface to a height equal to the radius

$R$ of the earth is

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

Explanation

The object is at earth surface initially.

Initial potential energy,

$P.E_i =\dfrac{-GMm}{R}$ where,

$GM = gR^2$

$\Rightarrow$

$P.E_i = -mgR$

Final Potential energy,

$P.E_f = \dfrac{-GMm}{2R}= -\dfrac{mgR}{2}$

$\therefore$ Change in potential energy,

$\Delta P.E = -\dfrac{mgR}{2} - (-mgR) = \dfrac{mgR}{2}$

A body of mass m is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

0%
$3 mg R$
0%
$\dfrac {1}{3} mg R$
0%
$2 mg R$
0%
$\dfrac {2}{3} mg R$

A body is taken to a height of

$nR$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is

0%
$(n\, +\, 1)^2$
0%
$(n\, +\, 1)^{-2}$
0%
$(n\, +\, 1)^{-1}$
0%
$(n\, +\, 1)$

Explanation

Acceleration due to gravity at the surface of the earth=

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

where

$R$ is the radius of earth.

Acceleration due to gravity at height

$nR=g'=\dfrac{GM}{(R+nR)^2}=\dfrac{GM}{R^2}\dfrac{1}{(n+1)^2}$

$\implies \dfrac{g}{g'}=(n+1)^2$

$g$ is the acceleration due to gravity on the earth's surface, the gain of the potential energy of an object of mass

$m$ raised from the surface of the earth to a height equal to the radius

$R$ of the earth will be :

0%
$2mgR$
0%
$mgR$
0%
$\dfrac{1}{2}mgR$
0%
$\dfrac{1}{4}mgR$

Explanation

The potential energy of an object at the surface of the earth

${ U }_{ 1 }=-\dfrac { GMm }{ R }$ .......(i)
The potential energy of the object at a height

$h = R$ from the surface of the earth

${ U }_{ 2 }=-\dfrac { GMm }{ R+h } =-\dfrac { GMm }{ R+R }$ .....(ii)
Hence, the gain in potential energy of the object

$\Delta U={ U }_{ 2 }-{ U }_{ 1 }$

$\Delta U=-\dfrac { GMm }{ R+R } +\dfrac { GMm }{ R }$

$\Delta U=-\dfrac { GMm }{ 2R } +\dfrac { GMm }{ R }$

$\Delta U=\dfrac { 1 }{ 2 } \dfrac { GMm }{ R }$
But we know that

$GM=g{ R }^{ 2 }$
Hence,

$\Delta U=\dfrac { 1 }{ 2 } \dfrac { g{ R }^{ 2 }m }{ R }$
or

$\Delta U=\dfrac { 1 }{ 2 } gRm$
or

$\Delta U=\dfrac { 1 }{ 2 } mgR$

Calculate angular velocity of earth so that acceleration due to gravity at

$60^o$ latitude becomes zero. (Radius of earth

$=6400$ km, gravitational acceleration at poles

$=10m{/s^2}, \cos 60^o=0.5$ )

0%
$7.8\times 10^{-2}rad/s$
0%
$0.5\times 10^{-3}rad/s$
0%
$1\times 10^{-3}rad/s$
0%
$2.5\times 10^{-3}rad/s$

Explanation

We have,

${ g }_{ 1 }=g-{ \omega }^{ 2 }R\cos ^{ 2 }{ \phi }$

According to the question,

$\omega=\dfrac{\pi}{3}$ and

$g_{1}=0$ .

So, we must have

$\omega =\sqrt { \dfrac { 4g }{ R } } =2.5\times { 10 }^{ -3 }rad/s$

Assuming

${ g }_{ \left( moon \right) }=\left( \dfrac { 1 }{ 6 } \right) { g }_{ earth }$ and

${ D }_{ \left( moon \right) }=\left( \dfrac { 1 }{ 4 } \right) { D }_{ earth }$ where

$g$ and

$D$ are the acceleration due to gravity and diameter respectively, the escape velocity from the moon is:

0%
$\dfrac { 11.2 }{ 24 }\ { kms }^{ -1 }$
0%
$11.2\times \sqrt { 24 } \ { kms }^{ -1 }$
0%
$\dfrac { 11.2 }{ \sqrt { 24 } }\ { kms }^{ -1 }$
0%
$11.2\times 24\ { kms }^{ -1 }$

Explanation

Escape velocity,

${ v }_{ e }=\sqrt { 2gR }$

$\Rightarrow \dfrac { { v }_{ m } }{ { v }_{ e } } =\sqrt { \dfrac { \dfrac { 1 }{ 6 } { g }_{ e }\times \dfrac { 1 }{ 4 } { R }_{ e } }{ { g }_{ e }\times { R }_{ e } } } =\dfrac { 1 }{ \sqrt { 24 } }$

$\Rightarrow { v }_{ m }=\dfrac { 11.2 }{ \sqrt { 24 } } \ { kms }^{ -1 }$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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