$$6.7 \times 10^{-11} \ ms^{-2}$$

MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 12

The gravitational potential energy of an object is due to

0%
its mass
0%
its acceleration due to gravity
0%
its height above the earth surface
0%
all of the above

Explanation

The gravitational potential energy of an object is given by:

$mgH$

(where

$m$ is the mass of an object ,

$g$ is its acceleration due to gravity and

$H$ is its height above the earth surface).

How much above the surface of earth does the acceleration due to gravity reduce by 36% of its value on the surface of earth? (radius of earth 6400 km)

0%
$800$ km
0%
$1600$ km
0%
$3200$ km
0%
$4266$ km

The value of universal gravitational constant

$'G'$ was first experimentally determined by :

0%
Galileo
0%
Newton
0%
Cavendish
0%
Kelvin

Explanation

Cavendish experiment . The Cavendish experiment , performed in 1797–1798 by British scientist Henry Cavendish, was the first experiment to measure the force of gravity between masses in the laboratory and the first to yield accurate value for the gravitational constant

Hence,

option

$C$ is correct answer.

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

$W_1 , W_2$ &

$W_3$ respectively, then :-

0%
$W_1 < W_2 > W_3$
0%
$W_1 = W_2 = W_3$
0%
$W_1 < W_2 < W_3$
0%
$W_1 > W_2 > W_3$

A wooden cylinder is floating on water in a beaker which is placed in a lift. When the lift is at rest, one third of the volume of the wooden cylinder is exposed above water. the lift now moves up with an acceleration g/The fraction of the volume exposed now is :

0%
1/3
0%
1/2
0%
1/6
0%
2/3

Let

$g$ be the acceleration due to gravity at earth's surface and

$K$ be the rotational kinetic energy of the earth Suppose the earth's radius decreases by

$2$ % keeping all other quantities same, then

0%
$g$ increases by $2$ % and $K$ increases by $2$ %
0%
$g$ increases by $4$ % and $K$ increases by $4$ %
0%
$g$ increases by $4$ % and $K$ increases by $2$ %
0%
$g$ increases by $2$ % and $K$ increases by $4$ %

A body weight W newton at the surface of the each, In weight at a height at a equal to half the radius of the earth will be:

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

A tunnel is dug along the diameter of the earth. There is a particle of mass m at the centre of the tunnel. Find the minimum velocity given to the particle so that just reaches to the surface of the earth. (R=radius of earth)

0%
$\sqrt{\dfrac{G M}{R}}$
0%
$\sqrt{\dfrac{G M}{2 R}}$
0%
$\sqrt{\dfrac{2 G M}{R}}$
0%
None of the above

Explanation

By conservation of energy,

Gain in K.E

$=$ Change in potential energy

$\dfrac 12 mv^2=m[-\dfrac{GM}{R}-[-\dfrac{3GM}{2R}]]$

A gravitational potential at centre

$=-\dfrac 32 \dfrac{GM}{R}$

Gravitational potential at surface

$=-\dfrac{GM}{R}$

Therefore,

$\dfrac 12mv^2=-\dfrac{GMm}{R}+\dfrac 32\dfrac{GMm}{R}$

$=\dfrac 12 mv^2=\dfrac{GMm}{2R}$

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

Body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If

$R$ is radius of the earth, the maximum height attained by the body is :-

0%
$\dfrac{R}{6}$
0%
$\dfrac{R}{3}$
0%
$\dfrac{2}{3}R$
0%
$R$

The gravitational potential at the centre of a square of side a when four point masses m each are kept at its vertices will be

0%
$4 \sqrt{2} \dfrac{Gm}{a}$
0%
$-4 \sqrt{2} \dfrac{Gm}{a}$
0%
$2 \sqrt{2} \dfrac{Gm}{a}$
0%
$-2 \sqrt{2} \dfrac{Gm}{a}$

Explanation

$\begin{array}{l} AO=\dfrac { { \sqrt { 2 } a } }{ 2 } =\dfrac { a }{ { \sqrt { 2 } } } \\ V=\dfrac { { -Gm } }{ r } \\ { V_{ net } }=\dfrac { { -4Gm } }{ r } =\dfrac { { -4Gm } }{ { \dfrac { a }{ { \sqrt { 2 } } } } } =\dfrac { { -4\sqrt { 2 } Gm } }{ a } \end{array}$ .

$\therefore$ Option

$B$ is correct.

1208969_1488707_ans_68a9e3b3ec734974b33cb2a5d423c288.JPG

Velocity of water in a river is

0%
Same everywhere
0%
More in middle and less near its bank
0%
Less in middle and more at banks
0%
Increases from one bank to other

The gravitational force between two point masses

${ m }_{ 1 }$ and

${ m }_{ 2}$ at separation r is given by

$F=k\dfrac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } .$
The constant k depends on :

0%
Depends on system of units only
0%
Depends on medium between masses only
0%
Depends on both (a) and (b)
0%
is independent of both (a) and (b)

The weight of a body at a distance 2R from the centre of earth of radius R is 2.5 N. If the distance is 3R from the centre of earth the weight of the body will be --------

0%
$1.1 N$
0%
$2.1 N$
0%
$3.1 N$
0%
$4.1 N$

How much percentage change in acceleration due gravity on an object occurs if object is initially at '2R' distance above the surface of the earth and falls through 'R' distance? (Here, R is the radius of earth)

0%
decrease by 300%
0%
increase by 300%
0%
decrease by 125%
0%
increase by 125%

Calculate the escape velocity for an atmospheric particle 9600 km above the earth 's surface , given acceleration due to gravity on the surface of earth is 10

$ms^{ -2 }$ .

0%
7.2 km/sec.
0%
8.6 km/sec.
0%
9.8 km/sec.
0%
6.4 km/sec.

The radius of the Earth shrinks by

$1$ %, its mass remaining the same. The percentage change in the value of

$g$ on surface of earth is

0%
$-2$ %
0%
$+2$ %
0%
$-3$ %
0%
$+4$ %

IF the acceleration due to gravity,

$g$ , is 10m/

${ s }^{ 2 }$ at the surface of the earth (radius

$6400 km$ ), then at a height of

$1600 m$ the value of

$g$ will be (in m/

${ s }^{ 2 }$ )

0%
$9.995$
0%
$5$
0%
$7.5$
0%
$2.5$

An aeroplane is moving with constant horizontal velocity u at height h. The velocity of a packet dropped from aeroplane, when it reaches on the earth's surface will be (g is acceleration due to gravity)

0%
$\sqrt { { u }^{ 2 }+2gh }$
0%
$\sqrt { 2gh }$
0%
$2gh$
0%
$\sqrt { { u }^{ 2 }-2gh }$

Explanation

Here, since there are no external forces and no non conservative forces, we can conserve the total mechanical energy of the system in the initial and final cases.

Assuming the mass of the packet to be

$m$ ,

For when the aeroplane is carrying the packet.

$TME_i = KE_i+PE_i$

$TME_i = \dfrac{1}{2}mu^2 + mgh$

When the packet reaches the ground, all the potential energy is converted into its kinetic energy. (

$PE_f = 0$ )

$TME_f = KE_f + PE_f$

$TME_f = \dfrac{1}{2}mv^2$

SInce,

$TME_i = TME_f$

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 + mgh$

$\dfrac{1}{2}v^2 = \dfrac{1}{2}u^2 + gh$

$v^2 = u^2+2gh$

$v = \sqrt{u^2+2gh}$

If the earth were to cases rotating about its own axis. The increase in the value of g in CGS system at a place of latitude of

${ 45 }^{ o }$ will be

0%
$2.68$
0%
$1.69$
0%
$3.36$
0%
$0.34$

A bird weighing

$1kg$ is sitting on the base of cage weighing

$1.5kg$ . The bird strarts flying inside the cage the weight of the bird cage assembly will now be:

0%
infinite
0%
$2.5kg$
0%
$3.5$
0%
$1kg$

If the radius of the earth is 6400 km and g = 10

${ m/s }^{ 2 }$ , the escape velocity from the earth is

0%
$8$ $km/s$
0%
$8\sqrt { 2 }$ $km/s$
0%
$8/\sqrt { 2 }$ $km/s$
0%
None of these

If a ball is projected with a velocity equal to 1/4th of the escape velocity from the surface of the earth. the height it will attain is ___________ [ R : Radius of Earth]

0%
$\dfrac{R}{4}$
0%
$\dfrac{R}{32}$
0%
$4R$
0%
$\dfrac{R}{16}$

Explanation

from newton's eqn. of motion

$h = \dfrac{v^2}{2g}$

$v = v_{esc} = \sqrt{2gR}$
as per question

$h = \dfrac{\left(\dfrac{1}{4}\sqrt{2gR}\right)^2}{2g} = \dfrac{2gR}{32g}$

$h = \dfrac{R}{16}$

Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will.

0%
Will become statioanry
0%
Keep floating at the same distance between them
0%
Move away from each other
0%
Move towards from each other

A body of mass

$m$ is raised from the earths, surface to an altitude equal to the radius (R) of the Earth. If

$g$ is the gravitational acceleration on the Earth's surface

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

Weight of a body at the equator of a planet is half of that at the poles. If peripheral velocity of a point on the equator of this planet is

$v_0$ , what is the escape velocity of a polar particle?

0%
$v_0$
0%
$2v_0$
0%
$3v_0$
0%
$4v_0$

Find the changes is gravitational potential energy if a particle of mass

$2\ kg$ is taken from the origin to the point

$(12\ m, 5\ m)$

0%
$-225\ J$
0%
$-240\ J$
0%
$-245\ J$
0%
$-250\ J$

Explanation

$\Delta v=+\displaystyle \int^{(0,0)}_{(12\ m,5\ m)}Edr =+ \displaystyle \int^{(0,0)}_{(12\ m,5\ m)} (5\hat i +12\hat j)2dr$

$=+(10\hat i +24\hat j) |r|_{(12\ m, 5\ m)}^{(0,0)}$

$=-(10\hat i +24\hat j) (12\hat i +5\hat j) =-240\ J$

Consider a satellite moving in a circular orbit around Earth. If K and V denote its kinetic energy and potential energy respectively, then(Choose the convention, where

$V=0$ as

$r\rightarrow \infty$ )

0%
$K=V$
0%
$K=2V$
0%
$V=2K$
0%
$K=-2V$
0%
$V=-2K$

Explanation

When a satellite moves in a circular orbit around the earth its

(i) Potential energy,

$\because V=-\dfrac{GMm}{r}$

(ii) Kinetic energy,

$K=\dfrac{1}{2}mv^2=\dfrac{GMm}{2r}\left[\because v=\sqrt{\dfrac{GM}{m}}\right]$

$\therefore v=-2K$ .

Find the change in potential energy if the particle is taken from

$(12\ m,0)$ to

$(0,5\ m)$

0%
$-10\ j$
0%
$-50\ J$
0%
zero
0%
$-60\ J$

Explanation

$\Delta v=+\displaystyle \int_{(0,5\ m)}^{(12\ m, 0)}Edr$

$=E \displaystyle \int^{5\hat i}_{5\hat j}dr=(10\hat i +24\hat j) [12\hat i -5\hat j]=0$

A man of mass

$m$ , standing at the bottom of the staircase, of height

$L$ climbs it and stands at its top.

0%
Work done by all forces on man is equal to the rise in potential energy $mgI$ .....
0%
Work done by all forces on man is zero.
0%
Work done by the gravitational force on man is $mgI$
0%
The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

Explanation

Work Done by gravitational force on man is

$(-mgL)$ as gravitational force is downward and displacement

$L$ is upward. The Work Done by man lift him up by muscular force will be

$(+mgL)$ as force applied by muscles is in the direction is displacement. So net Work Done

$=-mgL+mgL=0$ . Verifies option

$(b)$
As there is no displacement point where the reaction acts so, Work Done by reaction force is zero. As teh velocity of person atmost zero at top. So

$KE=0$ . Hence, Work Done by reaction force is zero. Verifies the option

$(d)$ .

Which of the following option are correct ?

0%
Acceleration due to gravity decreases with increasing altitude.
0%
Acceleration due to gravity increases with increasing depth ( assume the earth to be a sphere of uniform denisty)
0%
Acceleration due to gravity increases with increasing latitude.
0%
Acceleration due to gravity is independent of the mass of the earth.

Explanation

(a) The acceleration due to gravity at an altitude (height)

$gh = g ( 1 - \dfrac { 2h}{R})$
Increasing height

$(h)$ decreases the value of

$gh$ option (a) is correct.

(b) Assuming the earth to be a sphere of uniform density. the acceleration due to gravity at a particular depth

$(d) . gd = g( 1- \dfrac {d}{R})$ .
increasing depth

$(d)$ decreases the value of

$gd$ option (b) is incorrect.

$\lambda$ is latitude on earth then

$g \lambda = g = \omega^2 R cos^2 \lambda$
As value of cos

$\lambda$ decrease from

$0^0$ to

$90^0$ (from 1 to 0 ) the acceleration due to gravity increases from equator

$( \lambda = 0^0 )$ to pole

$( \lambda = 90^0 )$ option (c) is correct.

(d) The acceleration due to gravity on surface of earth is

$g= \dfrac {GM_{\theta} }{R^2_{ \theta } }$
So

$g$ on earth depends on mass of earth . option (d) is incorrect.

If the sun and the planets carried huge amounts of opposite charges,

0%
All three of kepler's law would still be valid
0%
Only the third law will be valid
0%
the second law will not change
0%
the first law will still be valid

The value of G is

0%
$9.8 \ N m^{2} kg^{-2}$
0%
$6.7 \times 10^{-11} \ Nm^{2}kg^{-2}$
0%
$6.7 \times 10^{-11} \ ms^{-2}$
0%
$6.7 \ Nkg^{-1}$

Explanation

Universal gravitational constant is represented as 'G'. Its value is

$G = 6.7 \times 10^{-11} \ Nm^{2}kg^{-2}$

At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on the surface ( R = radius of earth)

0%
$2R$
0%
$R$
0%
$1.44 \ R$
0%
$0.414 \ R$

Explanation

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

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

$\implies R+h= \sqrt{2} R$

$\implies h= ( \sqrt{2}-1) R= 0.414 \ R$

The depth at which the effective value of acceleration due to gravity is

$\dfrac{g}{4}$ is

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

Explanation

Value of g at depth 'd'

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

$g'=\dfrac {g}4$

$\implies \dfrac{g}{4} = g\bigg(1-\dfrac{d}{R} \bigg)$

$\implies d= \dfrac 34 R$

Acceleration due to gravity is '

$g$ ' on the surface of the earth. The value of acceleration due to gravity at a height of

$32 \ km$ above earth's surface is (Radius of the earth

$=6400 \ km$ )

0%
$0.9 \ g$
0%
$0.99 \ g$
0%
$0.8 \ g$
0%
$1.01 \ g$

Explanation

$h= 32 \ km, R=6400 \ km,$ so

$h<<R$

$g'= g\bigg( 1- \dfrac{2h}{R} \bigg) = g \bigg(1- \dfrac{2 \times 32}{6400} \bigg)$

$\implies g'= \dfrac{99}{100} g= 0.99 \ g$

The acceleration of a body due to the attraction of the earth (radius

$R$ ) at a distance

$2R$ from the surface of the earth is (

$g=$ acceleration due to gravity at the surface of the earth)

0%
$\dfrac{g}{9}$
0%
$\dfrac{g}{3}$
0%
$\dfrac{g}{4}$
0%
$g$

Explanation

Value of g at hight h,

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

h=2R (Given)

$g'= g \bigg( \dfrac{R}{R+2R} \bigg)^2 = \dfrac{g}{9}$

Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine

$100 \ km$ below the earth's surface (Given

$R= 6400 \ km$ )

0%
$9.66 \ m/s^2$
0%
$7.64 \ m/s^2$
0%
$5.06 \ m/s$
0%
$3.10 \ m/s^2$

Explanation

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

On putting above value.

$g' = 9.8 \bigg( 1- \dfrac{100}{6400} \bigg) = 9.66 \ m/s^2$

Kepler discovered

0%
Laws of motion
0%
Laws of rotational motion
0%
Laws of planetory motion
0%
Laws of curvilinear motion

The escape velocity on the moon is less than that on the earth.

0%
True
0%
False

Explanation

Escape velocity on a planet is given as :

$V_{esc}=\sqrt{\dfrac{2GM}{R}}=\sqrt{2gR}$
Since, acceleration due to gravity on moon is one-sixth of that on Earth, thus the escape velocity on the Moon is lees than that on the Earth. Hence, the statement is true.

At the center of the earth:

0%
the mass of a body is zero but the weight is not zero
0%
the mass of a body is not zero but the weight is zero
0%
both mass and weight of a body are zero
0%
both mass and weight of a body are not zero

Universal gravitational constant, G depends:

0%
on the nature of the particle
0%
on the medium present between the particles
0%
on time
0%
does not depend on any of these factors.

Explanation

Universal gravitational constant,

$G$ is independent of the nature of the particle, medium between the particles, and time. Its value is constant anywhere in the Universe, and hence it's called 'Universal'.

Practically the value of G for the first time was measured by ...........

0%
Newton
0%
Cavendish
0%
Archimedes
0%
Galileo

Explanation

The Cavendish has performed an experiment to calculate the gravitation force between two masses. and he gave the value of gravitational constant

$G$ , which is equal to

$6.75\times 10^{-11}\ \ Nm^2kg^{-2}$

An object moves from earths surface to the surface of the moon. The acceleration due to gravity on the earths surface is

$10 m/s^{2}$ . Considering the acceleration due to gravity on the moon to be

$1/6th$ times of that of earth. If

$R$ be the earths radius and its weight be

$W$ and the distance between the earth and the moon is

$D$ . The correct variation of the weight

$W'$ versus distance

$d$ for a body when it moves from the earth to the moon is

Explanation

At the earths surface the weight of body is,

$W = mg$
At the moon the distance of body from the earth's surface is

$d = D$
At moons surface the value of acceleration due to gravity is

$g' = \dfrac {g}{6}$

$\Rightarrow W' = \dfrac {W}{6}$
From the relation of acceleration due to gravity at height

$d$ above earth's surface,

$g' = \dfrac {gR^{2}}{(R + d)^{2}} ..... (i)$
For

$d = 0, \Rightarrow g' = g$ or

$W' = W$
At

$d = R$ ,

$\Rightarrow g' = \dfrac {gR^{2}}{4R^{2}} = \dfrac {g}{4}$
Or,

$W' = \dfrac {mg}{4} = \dfrac {W}{4}$
At distance

$d = D$ , when the body reaches surface of moon

$W' = \dfrac {mg}{6} = \dfrac {W}{6}$
Since, Eq. (i) suggests that variation of

$g$ with distance

$(d)$ is non-linear, hence the graph of

$W'$ vs

$d$ will be non-linear as well. Also

$d\uparrow, W'\downarrow$
Thus the correct variation is represented by following graph.
2113167_2005168_ans_1036a4729f7b4a5e8b946e35a40eb4e6.png

2113167_2005168_ans_1036a4729f7b4a5e8b946e35a40eb4e6.png

A boy can jump to a height

$h$ from ground on earth . What should be the radius of a sphere of density

$\delta$ such that on jumping on it, he escapes out of the gravitational field of the sphere?

0%
$\sqrt{\dfrac{4\pi G\delta }{3gh}}$
0%
$\sqrt{\dfrac{4\pi gh }{3G\delta }}$
0%
$\sqrt{\dfrac{3 gh }{4 \pi G\delta }}$
0%
$\sqrt{\dfrac{3 G\delta }{4 \pi gh }}$

Explanation

While jumping he converts his initial

$kinetic$

$energy$ to

$potential$

$energy$

$\dfrac{mv^2}{2}=mgh$

$v=\sqrt{2gh}$

Now this

$velocity$ has to be equal to

$escape$

$velocity$ on the sphere

$Escape$

$velocity=\sqrt{\dfrac{2GM_s}{R}}=\sqrt{\dfrac{8 \pi G R_s^2 \delta}{3}}$

By comparing both the velocities we get

$Radius(R_s)=\sqrt{\dfrac{3gh}{4 \pi G \delta}}$

At what distance above the surface of earth, the gravitational force will be reduced by

$10\%$ , if the radius of earth is

$6370$ Km.

0%
$750 Km$
0%
$650 Km$
0%
$450 Km$
0%
$344 Km$

Explanation

Gravitational force on the earth surface,

$F_{G_s}= \dfrac{GMm}{R^{2}}$

Gravitational force at a height 'h' above,

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

G.T. $F_{G_{h}}= \dfrac{90}{100}(F_{G_{S}})$

$\Rightarrow \dfrac{GMm}{(R+h)^{2}}=\dfrac{9}{10}(\dfrac{GMm}{R^{2}})$

$\Rightarrow R+h=\sqrt{\dfrac{10}{9}}(R)$

$\Rightarrow 6370+h= \dfrac{3.162}{3} (6370)$

$\Rightarrow h=\dfrac{0.162 \times 6370}{3}=344$ km

The value of

$g$ at a height

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

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

$d$ and

$h$ are much smaller than the radius of earth, then which one of the following is correct

0%
$d= \dfrac{h}{2}$
0%
$d= \dfrac{3h}{2}$
0%
. $d = 2h$
0%
$d= h$

Explanation

The value of gravity at a height

$h$ above earth's surface is given by

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

$d$ is given by

$g_d=g\left(1-\dfrac {d}{R}\right)$
Since these two values are same, we can see that

$d=2h$

The work done in shifting a particle of mass

$m$ from the centre of earth to the surface of the earth is

0%
$-mgR$
0%
$\dfrac{1}{2} mgR$
0%
$zero$
0%
$mgR$

Explanation

Given,The mass of the particle be

$m$ ,

Initilly it is on the centre of the earth, so

The work done in moving a particle from center of earth to its surface

$\ W=\int_0^{R}Fdr$

$=\int_0^R(-\dfrac{G(\rho\dfrac{4}{3}\pi r^3)m}{r^2})dr$

$=\int_0^R(-\dfrac{4G\rho \pi mr}{3})dr$

$=\dfrac{GMm}{2R},$ we know that

$F=\dfrac{GMm}{R^2}=mg$

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

(Since

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

So, The work done in shifting a particle of mass from the centre of the earth to the surface of the earth is

$=\dfrac{mgR}{2}.$

A particle of mass

$10 gm$ is kept on the surface of a uniform sphere of mass

$100 kg$ and radius

$10 cm$ . Find the work done against the gravitational force between them, to take the particle far away from the sphere.

$\left ( G= 6.67\times10^{-11}Nm^{2}/Kg^{2} \right )$

0%
$3.33\times10^{-9}J$
0%
$3.33\times10^{-10}J$
0%
$6.67\times10^{-9}J$
0%
$6.67\times10^{-10}J$

Explanation

At infinity, the gravitational potential energy of the system will be

$0$ .

Initially, the particle is outside sphere and hence sphere will behave as point object at its centre,

Therefore,

${ E }_{ 0 }=-\dfrac { GMm }{ R } =-\dfrac { (6.67\times { 10 }^{ -11 })\times(100)\times(0.01) }{ 0.1 } =-6.67\times{ 10 }^{ -10 }J$
We know that,

$W=\Delta U=U_{f}-U_{i} \implies W=-U_{i}$

Thus, work done is

$6.67\times { 10 }^{ -10 }J$

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to

$\sqrt{3}m$ is.

0%
$\dfrac{G}{3}J$
0%
$\dfrac{G}{4}J$
0%
$\dfrac{G}{5}J$
0%
$\dfrac{G}{6}J$

Explanation

Potential of a ring,

$V=\dfrac{-GM}{y}$

where,

$y=\sqrt{x^2+r^2}$

So initial, energy is

$\dfrac{-5G\times 1}{5}$

Final energy is

$\dfrac{-5G\times 1}{6}$

So the difference

$=$ Final

$-$ Initial

$=\dfrac{G}{6}$

The work done by an external agent to shift a point mass from infinity to the centre of the earth is W. Then choose the correct relation.

0%
W $=$ 0
0%
W > 0
0%
W < 0
0%
W $\leq$ 0

Explanation

$W_{ext}=\Delta U=U_{f}-U_{i}=U_{f}-0=U_{f}$

$W_{ext}=\displaystyle \frac{-3GMm}{2R} \rightarrow W_{ext}<0$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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