MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 15

For a circular orbit if velocity of an object is greater than orbital velocity and less than escape velocity then orbit will be

0%
circular
0%
Parabolic
0%
Elliptical
0%
Hyperbolic

An artificial earth satellite is taken from a higher orbit to a lower orbit

$i.e.r_2 < r_1$ . In this process,the gravitational potential energy

0%
in increased
0%
does not change
0%
in decreased
0%
is doubled

A particle of mass

$m$ is released from a height R ( radius of earth) from the surface of earth. When it reaches the earth's surface, it enters a tunnel leading to its centre. Its speed at the centre is

0%
$\sqrt {gR}$
0%
$\sqrt {2gR}$
0%
$\sqrt {3gR}$
0%
$\sqrt {5gR}$

How much energy will be necessary for making a body of 500 Kg to escape from the earth?

$[ g = 10 m/s^2$ , radius of earth

$= 6.4 \times 10^6 m]$

0%
$6.4 \times 10^8 J$
0%
$3.2 \times 10^{10} J$
0%
$6.4 \times 10^{12} J$
0%
$3.2 \times 10^6 J$

The gravitational field due to a mass distribution is

$\vec {I} = \dfrac {k}{x^{3}}\hat {i}$ where

$k$ is a constant. Taking the gravitational potential to be zero at infinity, its value at a distance

$x= d$ is

0%
$\dfrac {k}{2d^{2}}$
0%
$\dfrac {k}{d^{2}}$
0%
$\dfrac {k}{2d}$
0%
$\dfrac {k}{d}$

The weight of a body of mass

$250\ kg$ when it is at a height of

$4$ times the radius of earth, from earth's surface is :

0%
$25\ N$
0%
$50\ N$
0%
$64\ N$
0%
$100\ N$

The magnitude of gravitational potential at a distance

$\frac {a}{2}$ from the center

0%
$\dfrac { - 15GM } { 4 a }$
0%
$\dfrac { - 4GM } { 9 a }$
0%
$\dfrac { -11GM } { 8 a }$
0%
$\dfrac { - 6GM } { 7 a }$

At what height

$h$ above the earth surface, the value of

$g$ becomes

$\dfrac{g}{2}$ ? (where

$R$ is the radius of the earth)

0%
$(\sqrt 2 - 1)R$
0%
$\sqrt 2 R$
0%
$(\sqrt 2 + 1)R$
0%
$R\sqrt 2$

A man in a balloon rising vertically with an acceleration of

$4.9m/sec^2$ releases a ball

$2$ sec after the balloon is let go from the ground. The greatest height above the ground reaches by the ball is

$(g=9.8m/sec^2)$

0%
$14.7m$
0%
$19.6m$
0%
$9.8m$
0%
$24.5m$

Explanation

$a=4.9㎨$

Initial velocity

$=0$

Velocity of balloon and ball after

$2$ sec

$v=u+at=9.8㎧$

Height at this point

$h=\cfrac { 1 }{ 2 } a{ t }^{ 2 }$

$=0.5\times 4.9\times 4$

$=9.8m$

The ball will attain further height

$=s$

$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=mgs$

$s=\cfrac { { v }^{ 2 } }{ 2g } =4.9m$

Greatest height that can be achieved

$H=h+s=9.8+4.9=14.7m$

$W_1,W_2$ and

$W_3$ represent the work done in moving a particle from A to B along three different paths 1,2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between

$W_1,W_2$ and

$W_3$ :

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

A body weighted 980 N on the surface of the earth. It weight half way down to the centre of the earth is (Assume earth to be of uniform density)

0%
490 N
0%
125 N
0%
1960 N
0%
Zero

A man weights 100 kgf on the surface of earth. At what height above the surface of earth his weight will be 50 kgf? [where R is radius of the earth]

0%
$\frac{R}{4}$
0%
$\frac{R}{2}$
0%
$(\sqrt3 - 1)R$
0%
$(\sqrt2 - 1)R$

A particle hanging from a massless spring stretch it by

$2 cm$ at earths surface. How much will the same particle stretch the spring at height

$2624 km$ from the surface of earth

$?$ (radius of earth 6400 km)

0%
$1 cm$
0%
$2 cm$
0%
$3 cm$
0%
$4 cm$

A satellite is orbiting around the earth. By what percent age should increase its velocity so as to enable it escape away from earth:

0%
$41.4\%$
0%
$50\%$
0%
$82.8\%$
0%
$100\%$

Explanation

We know that orbital velocity

${v}_{0}=\sqrt{gr}$

Escape velocity

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

Thus, the increase in velocity would be

$dv={v}_{e}-{v}_{0}=\sqrt{2gr}-\sqrt{gr}$

$\Rightarrow dv=\left(\sqrt{2}-1\right)\sqrt{gr}$

$\Rightarrow dv=0.414\sqrt{gr}$

Now, the percentage increase would be

$\% dv=\dfrac{{v}_{e}-{v}_{0}}{{v}_{0}}\times 100$

$\% dv=0.414\dfrac{\sqrt{gr}}{\sqrt{gr}}\times 100=41.4\%$

The acceleration due to gravity becomes

$(\frac{g}{2})$ where g=acceleration due to gravity on the surface of the earth at a height equal to

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

Who among the following did first give the experimental value of

$G$ ?

0%
Cavendish
0%
Capernicus
0%
Brook Taylor
0%
None of these

A body of mass 2 kg is thrown into space with escape velocity 11.2 km/s, the escape velocity for another body of mass 5 kg will be

0%
38 km/s
0%
22.4 km/s
0%
50.6 km/s
0%
11.2 km/s

The height at which the acceleration due to gravity becomes

$g/9$ (where

$g=$ the acceleration due to gravity on the surface of the earth) in terms of

$R$ , (the radius of the earth) is:-

0%
$\frac {R}{2}$
0%
$\sqrt 2 R$
0%
$2R$
0%
$\frac {R}{\sqrt 2}$

The height above the surface of the earth where acceleration due to gravity is

$1/64$ of its value at surface of the earth is approximately

0%
$45\times 10^6$ m
0%
$54\times 10^6$ m
0%
$102\times 10^6$ m
0%
$72\times 10^6$ m

If radius of earth contracted by 0 .1%, its mass remains same, the weight of the body at the earth surface will increases by

0%
0 . 1 %
0%
0 . 2%
0%
0 . 3 %
0%
remains same

The ratio of acceleration due to gravity at a depth

$h$ below the surface of earth and at a height

$h$ above the earth surface of earth (h << radius of earth ) is

0%
Is constant
0%
Increases linearly with h
0%
Increases parabolically with h
0%
Decreases

The ratio of acceleration due to gravity at a depth h below the surface of earth and at a height h above the surface of earth for h < < R (where R is radius of earth)

0%
is a constant
0%
increases linearly with h
0%
increases parabolically with h
0%
decreases

The value of

$g$ will be

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

$\left( R _ { e } = 6400 k m \right).$

0%
$6400{ km }$
0%
$57600{ km }$
0%
$2560{ km }$
0%
$64000 k m$

The linear velocity of a body at the equator of the earth will be about

$($ Radius of earth at equator

$=6.4 \times 10^{6} \mathrm{m} )$

0%
$465$ $\mathrm{m} / \mathrm{s}$
0%
$320$ $\mathrm{m} / \mathrm{s}$
0%
$510$ $\mathrm{m} / \mathrm{s}$
0%
$720$ $\mathrm{m} / \mathrm{s}$

A particle is released from a very large separation from a sphere of mass

$m$ and radius

$R$ . In which a tunnel is present. Find the ratio of speed acquired by the particle as it reaches the center of the sphere to that of as it reaches its surface.

0%
$\sqrt{2}$
0%
$\sqrt{\dfrac{1}{2}}$
0%
$\sqrt{\dfrac{3}{2}}$
0%
$\sqrt{\dfrac{2}{3}}$

At what height from the surface of earth the gravitation potential and the value of g are

$-5.4 \times 10^7 J/kg$ and

$6.0 m/s^2$ respectively?
take the radius of earth as 6400 km:

0%
$2600 km$
0%
$1600 km$
0%
$1400 km$
0%
$2000 km$

To double the orbital speed V and halve the angular velocity

$\omega$ , The centripetal acceleration of a revolving body:

0%
Is quadrupled
0%
Remains Unchanged
0%
Is halved
0%
Is doubled

If the radius of a planet shrinks by 2% while its mass remaining the unchanged, the acceleration due to gravity on the planet's surface would

0%
Remain unchanged
0%
First decreases and then increases
0%
Decreases
0%
increases

A particle would take a time

$t$ to move down a straight tube from the surface of earth (supposed to be a homogeneous sphere) to its centre. If gravity were to remain constant, then the time would be

$t$ . The ratio of

$t/t$ will be

0%
$\dfrac{ \pi }{ 2\sqrt { 2 } }$
0%
$\dfrac{ \pi }{ \sqrt { 2 } }$
0%
${ \sqrt { 2 }\pi }$
0%
$\dfrac{ \pi }{ \sqrt { 2 } }$

A satellite is revolving around Earth in a circular orbit whose radius is R.work done by gravitational force in one revolution is

0%
zero
0%
2 $\mathrm { mgR } ^ { 2 }$
0%
$\mathrm { mgR } ^ { 2 }$
0%
$\mathrm { mgR }$

The acceleration due to gravity at a height

$1 km$ above the earth is the same as at a depth

$d$ below the surface of earth. Then

0%
$d = 1 \mathrm { km }$
0%
$d = \dfrac { 3 } { 2 } k m$
0%
$d = 2 \mathrm { km }$
0%
$d = \dfrac { 1 } { 2 } k m$

The escape velocity from a planet is

${ v }_{ 0 }.$ The escape velocity from a planet having twice the radius but same density will be-

0%
$0.5{\ v }_{\ 0 }$
0%
${\ v }_{\ 0 }$
0%
$2 {\ v }_{\ 0 }$
0%
$4 {\ v }_{\ 0 }$

A man is standing on an international space station, which is orbiting earth at an altitude 520

$\mathrm { km }$ with a constant speed 7.6

$\mathrm { km } / \mathrm { s }$ . If the men's weight is 50

$\mathrm { kg }$ , his acceleration is

0%
76 $\mathrm { km } / \mathrm { s } ^ { 2 }$
0%
7.6 $m / s ^ { 2 }$
0%
8.4 $m / s ^ { 2 }$
0%
10 $m / s ^ { 2 }$

In the usual notation, the acceleration due to gravity at a height

$h$ from the surface of the earth is .........

0%
$g= \frac{GM}{R+h}$
0%
$g= \frac{GM}{\sqrt{R+h}}$
0%
$g= \frac{GM}{(R+h)^2}$
0%
$g= GM(R+h)^2$

The value of the acceleration due to gravity at a height from Earth's surface decreases by

$1\%$ . What will be the value of height from the surface?(Take the radius of earth =

$6400\ km$ )

0%
$32\ km$
0%
$24\ km$
0%
$14\ km$
0%
$40\ km$

The change in the value of

$^ { \prime } g ^ { \prime }$ at a height

$^ { \prime }$ h'above the surface of the earth is same as at a depth 'd'. If d and h are much smaller than the radius of earth, then which one of the following is correct?

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

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= ______R where R is the radius of the earth.

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Suppose the acceleration due to gravity at the earth's surface is

$10 \mathrm { m } / \mathrm { s } ^ { 2 }$ and at the surface of mars it is

$4.0 \mathrm { m } / \mathrm { s } ^ { 2 } .$ A

$60 \mathrm { kg }$ passenger goes from the earth to the mars in a spaceship moving with a constant velocity. Neglect all other object in the sky. Which part of the figure best represent the weight (net gravitational force) of the passenger as a function of time?

0%
$A$
0%
$B$
0%
$C$
0%
$D$

The acceleration due to gravity slightly below and above the surface of the earth (radius R) at distances d and h, respectively. varies with the distance the centre as

If the radius of the earth is half of its present value and its mass 1/

$8^{ th }$ of the present mass , the 'g' value would have been reduced to

0%
$\dfrac { 1 }{ 8 } g$
0%
$\dfrac { 1 }{ 2 } g$
0%
$\dfrac { 1 }{ 3 } g$
0%
g

The acceleration due to gravity at a height

$\frac { R }{ 20 }$ from the surface is

$9ms^{-2}$ . Its value at a point, at an equal distance below the surface of the earth is

0%
$8.563ms^{-2}$
0%
$8.832ms^{-2}$
0%
$9.5ms^{-2}$
0%
$9.242ms^{-2}$

If the mass of the earth is M, its radius is R and gravitational constant G, Then the speed required to be given to a projectile so that it does not return to earth is.

0%
$\sqrt {\dfrac{{{\text{GM}}}}{{\text{R}}}}$
0%
$\sqrt {\dfrac{{{\text{2GM}}}}{{\text{R}}}}$
0%
$\sqrt {\dfrac{{{\text{3GR}}}}{{\text{M}}}}$
0%
$\sqrt {{\text{GRM}}}$

Wt. of body at height equal to radius of earth

0%
weight on earth
0%
$\dfrac{4}{3}$ of wt. on earth
0%
Double that of wt. of earth
0%
$\dfrac{1}{4}$ on wt. on earth

The diameter of a planet is four times that of the earth and its mean density is equal to that of earth. If the acceleration due to gravity on the earth surface is

$9.8 m/s$ the acceleration due to gravity on the planet surface is

0%
$4.9 ms^{-2}$
0%
$9.8 ms^{-2}$
0%
$19.6 ms^{-2}$
0%
$39.2 ms^{-2}$

At what height from the surface of the earth will the value of acceleration due to gravity be reduced by

$36$ % from the value at the surface ? (Given radius of earth

$= 6400\, km$ )

0%
$1600\, km$
0%
$1200\, km$
0%
$1000\, km$
0%
$800\, km$

A stone weighs

$100N$ on the surface of the earth. The ratio of its weight at a height of half the radius of the earth to its weight at a depth of half the radius of the earth will be approximately

0%
$2.6$
0%
$2.2$
0%
$1.8$
0%
$0.9$

The body of masses

$'m'$ is raised through a height above the surface of the earth so that the increase in

$P.E.$ is

$mgR/5$ where

$'R'$ is the radius of the earth. The height to which the body is raised is

0%
$R/8$
0%
$R/5$
0%
$R/2$
0%
$R/4$

The mass of a balloon with its contents is

$1.5\, kg$ . It is descending with an acceleration equal to half that of acceleration due to gravity. If it is to go up with the same acceleration keeping the volume same, its mass should be decreased by :

0%
$1.2\, kg$
0%
$1\, kg$
0%
$0.75\, kg$
0%
$0.5\, kg$

If the earth suddenly contracted to half of its radius without any change in its mass and its shape, duration of the day instead of

$24$ hours will be :

0%
$24$ hours
0%
$12$ hours
0%
$6$ hours
0%
$8$ hours

The acceleration due to gravity at a height

$\cfrac {R}{20}$ from the surface is

$9m/s^2$ Its value at a point at an equal distance below the surface of the earth is

0%
$8.563 ms^{-2}$
0%
$8.832 ms^{-2}$
0%
$9.426 ms^{-2}$
0%
$9.242 ms^{-2}$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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