MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 1

A planet in a distant solar system is

$10$ times more massive than the earth and its radius is

$10$ times smaller. Given that the escape velocity from the earth is

$11 km$

$\mathrm{s}^{-1}$ , the escape velocity from the surface ofthe planet would be

0%
$110 km \mathrm{s}^{-1}$
0%
$0.11 km \mathrm{s}^{-1}$
0%
$1.1 km \mathrm{s}^{-1}$
0%
$11 km \mathrm{s}^{-1}$

Explanation

$Escape\ velocity= \sqrt { \dfrac { 2Gm }{ r } } \\ \dfrac { { v }_{ new } }{ { v }_{ earth } } = \sqrt { \dfrac { { m }_{ new }{ r }_{ earth } }{ { m }_{ earth }{ r }_{ new } } } = \sqrt { \dfrac { 10 }{ \dfrac { 1 }{ 10 } } }= 10\\ { v }_{ new }= 110\ km{ s }^{ -1 }$

The mass of a spaceship is

$1000\ kg$ . It is to be launched from the earths surface out into free space. The value of

$g$ and

$R$ (radius of earth) are

$10\ m/s$ and

$6400\ km$ respectively. The required energy for this work will be :

0%
$6.4\times 10^{11}\ J$
0%
$6.4\times 10^{8}\ J$
0%
$6.4\times 10^{9}\ J$
0%
$6.4\times 10^{10}\ J$

Explanation

$\displaystyle W= \frac{GMm}{r}$

$\displaystyle =gR^2 \times \frac{m}{R}= mgR$

$=1000 \times 10 \times 6400 \times 10^3$

$=6.4 \times 10^{10}$

If g is the acceleration due to gravity on the earth's surface, the gain in 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, is:

0%
$2\mathrm{m}\mathrm{g}\mathrm{R}$
0%
$\displaystyle \dfrac{1}{2}\mathrm{m}\mathrm{g}\mathrm{R}$
0%
$\displaystyle \dfrac{1}{4}\mathrm{m}\mathrm{g}\mathrm{R}$
0%
$\mathrm{m}\mathrm{g}\mathrm{R}$

Explanation

Acceleration due to gravity on earth's surface

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

Gain in potential energy=

$\dfrac{GMm}{R}-\dfrac{GMm}{2R}=\dfrac{GMm}{2R}=\dfrac{1}{2}mgR$

A body weighs

$72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

0%
16 N.
0%
28 N
0%
32 N
0%
72 N

Explanation

Hint: Use formula of acceleration due to gravity at height h

Explanation:

Weight of the body on earth is given 72 N which is product of mass and acceleration due to gravity.

So,

$mg = 72$ .

Acceleration due to gravity at height h is given by the formula,

$g’ = g \dfrac{g{R_E}^2}{{{(R}_E+\dfrac{R_E}{2})}^2}$ , where

$R_E$ is radius of earth.

$g’ = \dfrac{4}{9} g$

So, weight on height h will be,

$W’ = mg’ = 72 * \dfrac{4}{9} = 32 N$

$Answer:$

Hence, option C is the correct answer.

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,

0%
the acceleration of S is always directed towards the centre of the earth.
0%
the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.
0%
the total mechanical energy of S varies periodically with time.
0%
the linear momentum of S remains constant in magnitude.

A body of mass '

$m$ ' 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%
$2mgR$
0%
$\displaystyle \frac{1}{3}mgR$
0%
$3mgR$
0%
$\displaystyle \frac{2}{3}mgR$

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

0%
$150 N$
0%
$200 N$
0%
$250 N$
0%
$100 N$

Explanation

Acceleration due to gravity at a depth d from surface of earth

$g' = g \left(1 - \dfrac{d}{R} \right)$ ....(1)
Where g = acceleration due to gravity at earth's surface
Multiplying by mass 'm' on both sides of (1)

$\implies mg' = mg \left(1 - \dfrac{d}{R} \right) \, \,\, \, Here, \left(d = \dfrac{R}{2} \right)$

$= 200 \left(1 - \dfrac{R}{2 R} \right) = \dfrac{200}{2} = 100 N$
1259353_1618754_ans_a1ff2f8615a6439e8b4aa1a51f2a9c7e.png

The ratio of escape velocity at earth

$(v_e)$ to the escape velocity at a planet

$(v_p)$ whose radius and mean density are twice as that of earth is:

0%
$1:2$
0%
$1:2\sqrt 2$
0%
$1:4$
0%
$1:\sqrt 2$

Explanation

We know that:

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

Given that:

$R'=2R$ and

$\rho'=2\rho$

$\Rightarrow M=\dfrac{4}{3}\pi R^3\times \rho$

$\Rightarrow M'=\dfrac{4}{3}\pi(2R)^3\times 2\rho=16M$

$\Rightarrow v_p=\sqrt{\dfrac{2G(16M)}{2R}}=\sqrt{\dfrac{16GM}{R}}$

$\Rightarrow \dfrac{v_e}{v_p}=\sqrt{\dfrac{1}{8}}=\dfrac{1}{2\sqrt{2}}$

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

Explanation

Acceteration due to gravity at height h,

$g_n=g_0 (1-\dfrac{2h}{R})$

$h=1km$
Acceleration due to gravity at depth d,

$d_d = g_0\left(1-\dfrac{d}{R}\right)$

$g_h=g_d$

$g_0(1-\dfrac{2h}{R})=g_0(1-\dfrac{d}{R})$

$\rightarrow d=2h$

$= 2\times 1 km$

$d=2km$

${v}_{e}$ is escape velocity and

${v}_{o}$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :

0%
${v}_{o}={v}_{e}$
0%
${v}_{e}=\sqrt {2{v}_{o}}$
0%
${v}_{e}=\sqrt {2}{v}_{o}$
0%
${v}_{o}=\sqrt {2}{v}_{e}$

Explanation

${v}_{escape}=\sqrt {\cfrac{2GM}{R}}$

${v}_{orbital}=\sqrt {\cfrac{GM}{R}}$

${v}_{escape}=\sqrt {2}{v}_{o}$

A particle of mass

$m$ is thrown upwards from the surface of the earth, with a velocity

$u$ . The mass and the radius of the earth are, respectively,

$M$ and

$R$ .

$G$ is gravitational constant and

$g$ is acceleration due to gravity on the surface of the earth. The minimum value of

$u$ so that the particle does not return back to earth is

0%
$\sqrt{\displaystyle\frac{2GM}{R}}$
0%
$\sqrt{\displaystyle\frac{2GM}{{R}^{2}}}$
0%
$\sqrt{2g{R}^{2}}$
0%
$\sqrt{\displaystyle\frac{GM}{{R}^{2}}}$

Explanation

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

${V}_{e} = \sqrt{2gR} = \sqrt{2\displaystyle\frac{GM}{{R}^{2}}}R = \sqrt{\displaystyle\frac{2GM}{R}}$ .

Find out the correct relation for the dependance of change in acceleration due to gravity on the angle at the latitude due to rotation of earth?

0%
$dg\propto cos\phi$
0%
$dg\propto { cos }^{ 2 }\phi$
0%
$dg\propto { cos }^{ 3/2 }\phi$
0%
$dg\propto \dfrac { 1 }{ cos\phi }$

Explanation

We know that, the effective gravity

$g'$ at a latitude

$\phi$ is given as

$g'=g-{ \omega }^{ 2 }R{ (\cos { \phi } ) }^{ 2 }$
Thus,

$g - g' = dg={ \omega }^{ 2 }R{ (\cos { \phi } ) }^{ 2 }$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

Effective acceleration due to gravity

$g' = g - R w^2 sin^2 \lambda$ where

$\lambda$ is the lattitude angle

At equator

$\lambda = 90^o$

$\implies g'_e = g - Rw^2$

$(sin 90^o = 1 )$

At poles

$\lambda = 0^o$

$\implies g'_p = g$

$(sin 0^o = 0 )$

$\therefore$

$g'_p - g'_e = g - (g - Rw^2) = Rw^2$

$\implies$

$g'_p - g'_e \propto w^2$

Hence, Assertion is correct.

Acceleration due to gravity at the surface of earth

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

But

$R_p < R_e$

$\implies g_p > g_e$

Hence, the reason statement is also correct but it does not explain the assertion correctly.

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%
$\left (\dfrac {n}{n + 1}\right )mgR$
0%
$\left (\dfrac {n}{n - 1}\right )mgR$
0%
$nmgR$
0%
$\dfrac {mgR}{n}$

Explanation

Gravitational potential energy of mass

$m$ at any point at a distance

$r$ from the centre of earth is

$U = -\dfrac {GMm}{r}$
At the surface of earth

$r = R$ ,

$\therefore U_{s} = -\dfrac {GMm}{R} = -mgR \left (\because g = \dfrac {GM}{R^{2}}\right )$
At the height

$h= nR$ from the surface of earth

$r = R + h = R + nR = R (1 + n)$

$\therefore U_{h} = -\dfrac {GMm}{R(1 + n)} = -\dfrac {mgR}{(1 + n)}$
Change in gravitational potential energy is

$\triangle U = U_{h} - U_{s} = -\dfrac {mgR}{(1 + n)} - (-mgR)$

$=-\dfrac {mgR}{1 + n} + mgR$

$= mgR\left (1 - \dfrac {1}{1 - n}\right ) = mgR\left (\dfrac {n}{1 + n}\right )$ .

The escape velocity for a body projected vertically upwards from the surface of earth is

$11$ km/s. If the body is projected at an angle of

$45^{0}$ with the vertical, the escape velocity will be

0%
$11\sqrt{2}km/s$
0%
$22$ $km / s$
0%
$11$ $km / s$
0%
$11/\sqrt{2}$ $km / s$

Explanation

Escape speed of a body from Earth's surface is given by:

$v_{min}= \sqrt {2gR}$
This expression is obtained by conservation of energy and doesn't involve in which direction the body is thrown/projected.
So, irrespective of the angle of projection, escape speed of the body from Earth's surface remains constant i.e.

$\approx 11$ km/s

$g$ on the surface of the Earth is

$9.8\ ms^{-2}$ , its value at a height of

$6400 km$ is: (Radius of the Earth

$= 6400km$ )

0%
$4.9\ ms^{-2}$
0%
$9.8\ ms^{-2}$
0%
$2.45\ ms^{-2}$
0%
$19.6\ ms^{-2}$

Explanation

The expression for

$g$ is given by:

$g_R =\dfrac {GM}{R^2}$
The value of

$g$ changes according to

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

$\displaystyle \dfrac{g_{h}}{g_R} = \dfrac{R^{2}}{(R+h)^{2}}$
Here,

$h = R$
So,

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

$\Rightarrow \dfrac {g_h}{g_R}=(\dfrac {R}{R+R})^2$

$\Rightarrow g_h=g_R\times\dfrac {1}{4} = 9.8\times 0.25=2.45\ m/s^2$

A satellite is revolving around the earth in an elliptical orbit. Its speed will be

0%
same at all points of the orbit
0%
different at different point of the orbit
0%
maximum at the farthest point
0%
minimum at the nearest point

Which of the following quantities remain constant in a planetary motion, when seen from the surface of the sun?

0%
Kinetic energy
0%
Angular speed
0%
Speed
0%
Angular momentum

Explanation

When viewed from the inertial frame of the sun, the planets seem to execute elliptical orbital motion with changing (

$K.E$ ) (Kepler's first law) but constant angular momentum (due to the force being central).

The escape velocity from the earth for a rocket is 11.2 km/sec. Ignoring the air resistance, the escape velocity of 10 mg grain of sand from the earth will be

0%
0.112 km/sec
0%
11.2 km/sec
0%
1.12 km/sec
0%
None

The gravitational field is a conservative field. The work done in this field by moving an object from one point to another

0%
depends on the end-points only.
0%
depends on the path along which the object is moved.
0%
depends on the end-points as well as the path between the points.
0%
is not zero when the object is brought back to its initial position.

The escape velocity of a body depends upon its mass as

0%
$m^{0}$
0%
$m^{1}$
0%
$m^{3}$
0%
$m^{2}$

Explanation

Escape speed of a body from Earth's surface is given by:

$v_{min}= \sqrt {2gR}$
As we can see from the above equation, there is no 'mass' term. Implies it can be written as

$m^{0}$ .
So, the escape speed of a body is independent of its mass.

A gravitational field is present in a region. A point mass is shifted from

$A$ to

$B$ , along different paths shown in the figure. If

$W_{1}$ ,

$W_{2}$ and

$W_{3}$ represent the work done by gravitational force for respective paths, then

0%
$W_1=W_2= W_3$
0%
$W_1> W_2> W_3$
0%
$W_1> W_3> W_2$
0%
none of these

The earth retains its atmosphere. This is due to

0%
the special shape of the earth.
0%
the escape velocity, which is greater than the mean speed of atmospheric molecules.
0%
the escape velocity, which is less than the main speed of atmospheric molecules.
0%
the suns gravitational effect.

$g$ on the surface of the Earth is

$9.8$

$ms^{-2}$ , then it's value at a depth of

$3200$

$km$ (Radius of the earth

$= 6400$

$km$ ) is

0%
$9.8$ $ms^{-2}$
0%
$zero$
0%
$4.9$ $ms^{-2}$
0%
$2.45$ $ms^{-2}$

Explanation

The value of gravity changes as we move away from or towards the centre of the Earth.
This is given by:

$g_{R} =\dfrac{GM}{ R^{2} }$ , where

$M$ is the mass of a planet of radius

$R$
So,

$M=\dfrac{4}{3}\pi R^{3}\rho$ ; substituting in above equation

$\Rightarrow g_R = G \dfrac{4}{3} \pi \rho \times R$
Since we want the value of

$g$ at depth(

$d$ ) from the Earth's surface, we replace

$R$ by

$(R-d)$

$\Rightarrow g_d = G \dfrac{4}{3} \pi \rho \times (R-d)$

$\Rightarrow \dfrac{g_{R}}{ g_{d} } = \dfrac{R}{R-d}$

$\Rightarrow \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{d}{R})$

The given depth in the problem is

$d = 3200$ km, substituting we get,

$\dfrac{g_{d}}{ g_{R} } =(1- \dfrac{3200}{6400})$

${g_{d}}=9.8\times(1- \dfrac{3200}{6400})$

${g_{d}}=9.8/2$

${g_{d}}=4.9 m s^{-2}$

The value of quantity 'G' in the law of gravitation:

0%
depends on mass of earth only
0%
depends on radius of earth only
0%
depends on both mass and radius of earth
0%
is independent of mass and radius of the earth

The escape velocity from the earth for a rocket is

$11.2$ km/s ignoring air resistance. The escape velocity of

$10$ mg grain of sand from the earth will be

0%
$0.112$ km/s
0%
$11.2$ km/s
0%
$1.12$ km/s
0%
$0.0112$ kms $^{-1}$

The escape velocity of a sphere of mass

$m$ is given by

0%
$\sqrt{\dfrac{2GMm}{R_{e}}}$
0%
$\sqrt{\dfrac{2GM}{R_{e}^{2}}}$
0%
$\sqrt{\dfrac{2GMm}{R_{e}^{2}}}$
0%
$\sqrt{\dfrac{2GM}{R_{e}}}$

Explanation

For the body to escape Earth's gravitation and reach infinity, the initial kinetic energy has to be just higher than the Gravitational potential energy of the body on the surface.
Total energy of the body just after projecting is:

$E=\dfrac {1}{2}mv^2 - \dfrac {GMm}{R_e}$
Now this total energy has to be just greater than zero, so the minimum speed is achieved when kinetic energy is equal to gravitational potential energy.

$\Rightarrow \dfrac {1}{2}mv^2=\dfrac {GMm}{R_e}$

$\Rightarrow \dfrac {1}{2}v^2=\dfrac {GM}{R_e}$

$\Rightarrow v=\sqrt {\dfrac {2GM}{R_e}}$

The value of

$g$ at a height of

$100km$ from the surface of the Earth is nearly (Radius of the Earth

$=$ 6400km) (

$g$ on the surface of the Earth

$=$

$9.8m/s^{2}$ )

0%
$9.5 ms^{-2}$
0%
$8.5 ms^{-2}$
0%
$10.5 ms^{-2}$
0%
$9.8 ms^{-2}$

Explanation

Acceleration due to gravity changes with the height from Earth's surface as:

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

$\Rightarrow g^{'} = 9.8 (1-\dfrac {2\times 100}{6400})$

$\Rightarrow g^{'} = 9.8(1-\dfrac {1}{32})=9.49 m/s^2$

Planets rotate around the Sun in a path best described as

0%
elliptical
0%
circular
0%
parabola
0%
none of the above

The acceleration due to gravity at a depth of

$1600km$ inside the earth is

0%
$6.65$ ms $^{-2}$
0%
$7.35$ ms $^{-2}$
0%
$8.65$ ms $^{-2}$
0%
$4.35$ ms $^{-2}$

Explanation

The value of

$g$ at a depth

$d$ is given by:

$g^{'} = g \left ( 1-\dfrac{d}{R}\right)=9.81 \left(1-\dfrac {1600}{6400} \right) =9.81\times \left( \dfrac{3}{4} \right)=7.35$

$ms$

$^{-2}$

At what height, the value of '

$g$ ' is half that on the surface of the earth of radius

$R$ ?

0%
$R$
0%
$2R$
0%
$0.414R$
0%
$0.75R$

Explanation

We have,

$g=\dfrac {GM}{r^2}$ ;

$r$ is the distance from the centre of Earth.

$\Rightarrow g\propto \dfrac {1}{r^2}$ above the surface of the earth.

$\Rightarrow \dfrac {g_s}{g_h}=\dfrac {r_h^2}{r_s^2}$

Since we want the value of

$g_s$ to be half, we replace

$g_h$ by

$g_s/2$

$\Rightarrow \dfrac {g_s}{g_s/2}=[\dfrac {R+h}{R}]^2$

$\Rightarrow 2=[\dfrac {R+h}{R}]^2$

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

$\Rightarrow \dfrac {h}{R}=1.414-1=0.414$

$\Rightarrow h=0.414R$

The ratio of escape velocities of two planets if

$g$ values on the two planets are

$9.9$ m/s

$^{2}$ and

$3.3$ m/s

$^{2}$ and their radii are

$6400km$ and

$3400km$ respectively is

0%
$2.36 : 1$
0%
$1.36 : 1$
0%
$3.36 : 1$
0%
$4.36 : 1$

Explanation

Escape speed,

$v_e=\sqrt {2gR}$

$\Rightarrow \dfrac {v_2}{v_1}=\sqrt {\dfrac {g_2R_2}{g_1R_1}}=\sqrt {\dfrac {9.9\times6400}{3.3\times3400}}=2.37$

$\Rightarrow v_2:v_1=2.37:1$

The difference in

$PE$ of an object of mass

$10kg$ when it is taken from a height of

$6400km$ to

$12800km$ from the surface of the earth is

0%
$3.11\times 10^{8}$ J
0%
$1.565\times 10^{8}$ J
0%
$2.65\times 10^{8}$ J
0%
$4.5\times 10^{8}$ J

Explanation

Difference in potential energy is given by:

$PE=-GMm\left(\dfrac {1}{R_1}-\dfrac {1}{R_2}\right)$
But,

$R_1 = 6400$

$km$ and

$R_2 = 12800$

$km$

$\Rightarrow PE=-6.67\times10^{-11}\left(5.97\times10^{24}\right) (10) \left(\dfrac {1}{6400}-\dfrac {1}{12800}\right)\times10^{-3}$

$\Rightarrow PE = 0.00311\times10^{11}J$

The kinetic energy needed to project a body of mass

$m$ from earth's surface

$($ radius

$R$

$)$ to infinity is

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

Explanation

$E=\dfrac {1}{2}mv^2 - \dfrac {GMm}{R} = \dfrac {1}{2}mv^2 - mgR$
To escape Earth's gravitation, this energy has to be positive:

$\Rightarrow \dfrac {1}{2}mv^2 - mgR \geq 0 \Rightarrow K.E \geq mgR$
So, the minimum Kinetic energy required is

$mgR$ .

A spring balance is on sea level. If a body is weighed with this balance at consecutively increasing heights from earth's surface, the weight indicated by the balance:

0%
will go on decreasing continuously
0%
will go on increasing continuously
0%
will remain same
0%
will first increase and then decrease

Acceleration due to gravity becomes half at a depth of half the radius of the earth.

0%
True
0%
False

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

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

and

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

so when

$r=\dfrac{R}{2}$ then g will be half of g when r=R.

so the above statement is true.

An object is weighted in the following places using a spring balance. In which place will it weight the heaviest?

0%
On the Moon
0%
At the equator
0%
At the pole
0%
In outer space

Universal gravitational constant is indepndent of the intervening medium.

0%
True
0%
False

Explanation

Yes, universal gravitational constant is constant always, irrespective of the nature of medium and distance between two bodies.

When material between two bodies is changed then neither

$G$ is changed nor gravitational force between bodies.

So the above statement is true.

A satellite is orbiting around the earth. Then, the plane of the orbit:

0%
must be equatorial plane.
0%
must pass through the earth's centre.
0%
must pass through the poles.
0%
any plane with centre lying on the axis of rotation of the earth.

At what height in km over the earth's pole, the free fall acceleration decreases by one percent? (Assume the radius of the earth to be 6400 km).

0%
32
0%
64
0%
80
0%
1.253

Explanation

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

$0.99=\dfrac {1}{(1+\dfrac {h}{R})^2}$

$0.995=\dfrac {R}{R+h}$

$0.995h=0.0050R$

$h=32.2 km$

The value of g on the earth's surface is

$980 cm s^{-2}$ . Its value at a height of 64km from the earth's surface is:

0%
$960.40 cm s^{-2}$
0%
$984.90 cm s^{-2}$
0%
$982.45 cm s^{-2}$
0%
$977.55 cm s^{-2}$

Explanation

$g=980 cm s^{-2}$

$g_h=g(1-\dfrac {2h}{R})=980(1-\dfrac {2(64)}{6400})$

$=960.40 cm s^{-2}$

State whether the given statement is True or False :

The value of G is high if the radius of the body is more and less if radius is less.

0%
True
0%
False

The factor(s) affecting the value of 'g' is(are):

0%
depth of the body from earth's surface
0%
height of the body from the earth's surface
0%
mass of the body
0%
both A and B

In the relation F=

$\dfrac{G M m}{r^{2}}$ , the quantity

$G$

0%
depends on the value of g at the place of observation.
0%
is used only when the earth is one of the two masses.
0%
is greatest at the surface of the earth.
0%
is universal constant in nature.

Explanation

$G$ is the universal gravitational constant which remains constant at all places in the universe.

$G$ is equivalent to the force of attraction between two bodies of unit mass and unit distance apart.

The force of attraction between two unit point masses separated by a unit distance is called

0%
Gravitational potential
0%
Acceleration due to gravity.
0%
Gravitational field
0%
Universal gravitational constant.

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

0%
Decreases
0%
Increases
0%
Remains constant
0%
Data insufficient

Explanation

$g=(\dfrac{4\pi G\rho_o}{3})(1-\dfrac{d}{R})$ where

$\rho_o$ is the density of earth,

$R$ is the radius of Earth,

$d$ is depth
So as d increases acceleration due to gravity decreases

A body has a weight of

$10 kg$ on the surface of the Earth. What will be its mass and weight when taken to the centre of the Earth?

0%
$10$ kg, zero
0%
zero, zero
0%
$10$ kg, $10$ g
0%
zero, $10$ g

Explanation

Mass is constant and hence will be

$10 kg$ .
W=mg and

$g=\dfrac{4 \pi G \rho_0}{3}(1- \dfrac{d}{R})$ where R is the radius of earth,

$\rho_0$ is the density of earth, d is the depth.
At the center of earth

$d=R$ , and hence

$g=0$ and

$W=0$

The weight of an object at the centre of the earth of radius

$R$ is

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Zero
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Infinite
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$R$ times the weight at the surface of the earth.
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${1/R^2}$ times the weight at surface of the earth.

Explanation

Gravitational pull for any point below the earth's surface is given by

$g= \dfrac{4\pi G(R_e-d) \rho}{3}$
At center of earth depth

$d= R_e$
Therefore,

$g= \dfrac{4\pi G(R_e-R_e) \rho}{3}$ = 0

The minimum velocity of projection to go out from the earth's gravitational pull is called

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Terminal velocity
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Escape velocity
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Angular velocity
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Orbital velocity

State whether the given statement is True or False :

The value of G depends upon the mass of the two objects.

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True
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False

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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