MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 3

The value of gravitational constant depends upon :

0%
temperature of the atmosphere
0%
masses
0%
distance between the masses
0%
none of these

Explanation

The gravitational constant is a universal constant. It remains as it is in any condition.

It does not depend on any of the factors like the temperature, mass, distance between the masses.

Value of G is

$6.67 \times 10^{-11} \, Nm^2 kg^{-2}$

If R

$=$ radius of the earth and g

$=$ acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to

0%
$r$
0%
$r^{2}$
0%
$r^{-2}$
0%
$r^{-1}$

Explanation

For

$r>R$ , earth can be considered as a point mass.
Therefore,

$F = \dfrac{GMm}{r^2}$
From the equation,

$F$ is inversely proportional to

$r^2$ .

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 nearly

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

Explanation

Acceleration due to gravity a height

$h$ from earth's surface is given by:

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

Here,

$g_h = \dfrac{g}{2}$

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

$\Rightarrow \dfrac {R+h}{R}=1.414$

$\Rightarrow \dfrac {h}{R}=0.414$

$\Rightarrow h=0.414\times 6400=2649.6\ km$

A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

0%
positive
0%
negative
0%
zero
0%
may be positive or negative depending upon its initial velocity

Explanation

When a body is projected with an initial speed

$v_{i}$ from a point at a distance

$R_{E}+h$ from Earth's centre, the total energy is given by

$E = \dfrac {1}{2}m{v_{i}^{2}} - \dfrac {GmM_{E}}{(R_{E}+h)}$
The body escapes Earth's gravitation when the first term in the right hand side is equal or greater than the second term, that is the escape speed. When that speed is less than the escape speed, first term is smaller than the second term implies Gravitational potential energy is higher than the kinetic energy. Hence the total energy will be negative.

If R

$=$ radius of the earth and g

$=$ acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r<R) from the centre of the earth is proportional to

0%
$r$
0%
$r^{2}$
0%
$r^{-2}$
0%
$r^{-1}$

Explanation

$F(d) = GM'm/r^2$

$M' = Mr^3/R^3$

$F(d)= GMmr/R^3$

$r = R-d$

$g(d) = GM(R-d)/R^3$

$g(d) = g[(R-d)/R]$

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

Therefore, the gravitational field varies proportional to

$r$ .

A man covers

$60m$ distance in one minute on the surface of earth. The distance he will cover on the surface of moon in one minute is

$\left ( g_{m}= \dfrac{g_{e}}{6} \right )$

0%
$60 m$
0%
$60 X 6 m$
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$\dfrac{60}{6}m$
0%
$\sqrt{60}m$

The depth at which the value of

$g$ becomes

$25\%$ of that at the surface of the earth.

$($ Radius of the earth

$=$ 6400km.

$)$

0%
4800 km
0%
2400 km
0%
3600 km
0%
1200 km

Explanation

The value of

$g$ at a depth is given by:

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

Here,

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

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

$\Rightarrow \dfrac {d}{R}=\dfrac {3}{4}$

$\Rightarrow d = 0.75\times 6400 = 4800\ km$

A body has weight (W) on the ground. The work which must be done to lift it to a height equal to the radius of the earth, R, is

0%
equal to W $\times$ R
0%
greater than W $\times$ R
0%
less than W $\times$ R
0%
Cannot be determined

The ratio of the radius of a planet A to that of planet B is

$r$ . The ratio of accelerations due to gravity for the two planets is

$x$ . The ratio of the escape velocities from the two planets is :

0%
$\sqrt{rx}$
0%
$\sqrt{\dfrac{r}{x}}$
0%
$\sqrt{r}$
0%
$\sqrt{\dfrac{x}{r}}$

Explanation

Escape Velocity is given by:

$V_{e}=\sqrt{2gR}$
Given:

$\dfrac{R_A}{R_B} = r$ and

$\dfrac{g_A}{g_B} = x$
Taking ratio of the escape velocities of the

$2$ planets:

$\dfrac{V_A}{V_B} = \sqrt{rx}$

The ratio of acceleration due to gravity at a depth

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

$h$ above the surface for

$h<<R$

0%
is constant.
0%
increases linearly with h.
0%
varies parabolically with h.
0%
decreases.

Explanation

Acceleration due to gravity at a depth $h$ below the surface of earth is

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

Acceleration due to gravity at a height $h$ above the surface of earth is

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

where, $g$ is the acceleration due to gravity on the surface and $R$ is the radius of earth.

So, $\dfrac{g_d}{g_h} = \dfrac{(R - h)}{ (R - 2h)}$

Since $R>> h$

We can conclude that this ratio is constant.

Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by

0%
A goes on decreasing and that of B goes on increasing
0%
B goes on decreasing and that of A goes on increasing
0%
each decreases at the same rate
0%
each decreases at different rates

Explanation

The variation of 'g' with altitude is:

$g' = g(1- \dfrac{2h}{R})$
The variation of 'g' with depth is:

$g' = g(1- \dfrac{d}{R})$
So, from the above equations we can deduce that if a person goes deep into a mine or to a height above the Earth's surface, the gravitational field decreases at different rates.

The acceleration due to gravity on the surface of moon is

$1/6$ that on the earth and the diameter of the earth is

$4$ times the diameter of the moon. The rough ratio of the escape velocity of the moon to that of the earth is

0%
$1 : 4$
0%
$4 : 1$
0%
$5 : 1$
0%
$1 : 5$

Explanation

Acceleration due to gravity on the surface of a celestial body is given by:

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

It is given that

$g_e:g_m = 1:6$

and

$R_e:R_m = 4:1$

Escape velocity for a planet is given by

$v = \sqrt {2gR}$

Ratio of escape velocity is given by:

$\dfrac{v_m}{v_e} = \dfrac{\sqrt{2g_mR_m}}{\sqrt{2g_eR_e}}$

$=\dfrac{1}{\sqrt{6 \times 4}} \approx \dfrac{1}{5}$

If the radius of earth decreases by

$10\%$ , the mass remaining unchanged, then the acceleration due to gravity

0%
decreases by $19\%$
0%
increases by $19\%$
0%
decreases by more than $19\%$
0%
increases by more than $19\%$

Explanation

The acceleration due to gravity is given by:

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

$\Rightarrow g\propto \dfrac {1}{R^2}$

$\Rightarrow \dfrac {\Delta g}{g}=-2\dfrac {\Delta R}{R}=-2(-0.1)=0.2$
So, the value of

$g$ increases by

$20\%$ .

What is the escape velocity from the surface of the earth of radius

$R$ and density

$\rho$ ?

0%
$2R\sqrt{\dfrac{2\pi \rho G}{3}}$
0%
$2\sqrt{\dfrac{2\pi \rho G}{3}}$
0%
$2\pi \sqrt{\dfrac{R}{g}}$
0%
$\sqrt{\dfrac{2\pi G\rho }{R^{2}}}$

Explanation

Escape speed is given by

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

But

$M=Volume\times density \Rightarrow M=\dfrac{4}{3} \pi r^3 \rho$

$\Rightarrow v_e=\sqrt {2G\dfrac {\dfrac {4}{3}\pi R^3\rho }{R}}$

$\Rightarrow v_e=2R\sqrt {\dfrac {2\pi \rho G}{3}}$

The ratio of the radii of planets A and B is

$K_{1}$ and ratio of accelerations due to gravity on them is

$K_{2}$ The ratio of escape velocities from them will be:

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

Explanation

Escape speed is, $v_e=\sqrt {2gR}$
$\therefore \dfrac {v_1}{v_2}=\sqrt{\dfrac {g_1R_1}{g_2R_2}}$
It is given, $\dfrac {R_1}{R_2}=K_1$ and $\dfrac {g_1}{g_2}=K_2$
$\Rightarrow \dfrac {v_1}{v_2}=\sqrt {K_1K_2}$

A spaceship is launched into a circular orbit of radius

$R$ close to the surface of earth. The additional velocity to be imparted to the spaceship in the orbit to overcome the earth's gravitational pull is :

$($

$g =$ acceleration due to gravity

$)$

0%
$1.414Rg$
0%
$1.414\sqrt{Rg}$
0%
$0.414Rg$
0%
$0.414\sqrt{gR}$

Explanation

Velocity of the spaceship when in a circular orbit of radius

$R =\sqrt{gR}$
where g is the acceleration due to gravity and

$R$ is the radius of the earth .
Velocity needed to escape earth's gravitational pull

$=\sqrt{2gR}$

$\therefore$ Additional velocity to be imparted

$= \sqrt{2gR} - \sqrt{gR}$

$=(\sqrt{2} -1)\sqrt{gR}$

A particle hanging from a spring stretches it by

$1 cm$ at earth's surface. Radius of earth is

$6400 km$ . At a place

$800 km$ above the earths surface, the same particle will stretch the spring by

0%
$0.79 cm$
0%
$1.2 cm$
0%
$4 cm$
0%
$17 cm$

Explanation

as,

$mg=kx$

where,

$x=1cm, k=$ spring constant

for second case

$mg'=kx'$

Hence

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

$x'=xg'/g$

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

Earth's radius,

$R=6400$

At height,

$h=800$

$x'=\dfrac{1\times (6400 \times 10^3)^2}{(7200 \times 10^3)^2}$

$x'=0.79 cm$

If the radius of earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth's surface would

0%
decrease
0%
remain unchanged
0%
increase
0%
nothing will happen

Explanation

The acceleration due to gravity is given by:

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

$\Rightarrow g\propto \dfrac {1}{R^2}$

$\Rightarrow \dfrac {\Delta g}{g}=-2\dfrac {\Delta R}{R}=-2(-0.01)=0.02$
So, the value of

$g$ increases by 2 percent.

A tunnel is dug along a diameter of the Earth. The force on a particle of mass

$m$ placed in the tunnel at a distance

$x$ from the centre is:

0%
$\dfrac{GM_{e}m}{R^{3}}x$
0%
$\dfrac{GM_{e}m}{R^{2}}x$
0%
$\dfrac{GM_{e}m}{R^{3}x}$
0%
$\dfrac{GM_{e}mR^{3}}{x}$

Explanation

The value of

$g^{'}$ at depth

$d$ is given by,

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

The value of

$g^{'}$ at a distance

$x$ form Earth's centre is

$g^{'}= g \left( 1-\dfrac{R-x}{R} \right) = \dfrac{gx}{R}$

But

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

$\Rightarrow g^{'}=\dfrac{\left(\dfrac {GM}{R^2} \right)}{R} x=\dfrac {GMx}{R^3}$

Force on a particle of mass

$m$ is

$mg=\dfrac {GMmx}{R^3}$

A space craft is launched in a circular orbit very close to the earth. What additional velocity should be given to the space craft so that it might escape the earth's gravitational pull:

0%
$20.2 Kms^{-1}$
0%
$3.25 Kms^{-1}$
0%
$8 Kms^{-1}$
0%
$11.2 Kms^{-1}$

Explanation

Since the spacecraft is moving very close to the surface of earth, we assume its orbital radius to be R(the radius of the earth); implying its current orbital velocity is:

$V_{o}\, = \sqrt{gR}$ .
But by definition of escape velocity, we know that to escape the surface of earth, the speed required is

$V_{e}\, = \sqrt{2gR}$ .
Hence, the additional speed to be given to the spacecraft is

$V_{e}\, -\, V_{o}$ .
Using

$g=9.81$ and

$R=6400\ km$ , we get the required additional speed as

$3.25km\, s^{-1}$ .

$R$ is radius of the earth and

$W$ is work done in lifting a body from the ground to an altitude

$R$ , the work which should be done in lifting it further to twice that altitude is:

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

Explanation

Work done is equal to change in PE.

$W = - \Delta U = U_1-U_2 = -\dfrac{GMm}{R}-(- \dfrac{GMm}{2R}) = -\dfrac{GMm}{2R}$

Work done in lifting body from

$2R$ to

$4R$ .

$W' = - \Delta U = U_1-U_2 = -\dfrac{GMm}{2R}-(- \dfrac{GMm}{4R}) = -\dfrac{GMm}{4R} =\dfrac{W}{2}$

A body of mass

$m$ is raised from the surface of the earth to a height

$nR$ (

$R$ -radius of earth). Magnitude of the change in the gravitational potential energy of the body is (

$g$ - acceleration due to gravity on the surface of earth) :

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

Explanation

Change in P.E.

$= \dfrac {-GMm}{(n+1)R} - (\dfrac{-GMm}{R}) = (\dfrac {GMm}{(n+1)R}) ( -1 + (n+1))$

$= (\dfrac {GMm}{R}) \dfrac{(n)}{n+1}$
But

$g = GM/R^2$ or

$GM = g R^2$
Putting this value, we get Change in P.E.

$= mgR (\dfrac {n}{n+1})$

The work done to increase the radius of orbit of a satellite of mass

$m$ revolving around a planet of mass

$M$ from orbit of radius

$R$ into another orbit of radius

$3R$ is :

0%
$\dfrac{2GMm}{3R}$
0%
$\dfrac{GMm}{R}$
0%
$\dfrac{GMm}{6R}$
0%
$\dfrac{GMm}{24R}$

Explanation

The gravitational force '

$F$ ' on satellite of mass 'm' revolving around the planet in the orbit of radius '

$R$ ' is

$F = - \dfrac{GMm}{R^2}$
Where

$G$ is gravitational constant

$M$ is Mass of the planet.
The work done in increasing the radius of orbit of the satellite through '

$dr$ ' is

$W = F dr$
but work done is equal to change in gravitational potential '

$-dV$ '
therefore

$-dV = F dr$

$V = - \int_{R}^{3R}F dr$

$V = -\dfrac{GMm}{R^2}\int_{R}^{3R}-(F dr)$

$V = GMm\int_{R}^{3R}\dfrac {1}{R^2}dr$

$V = \dfrac{GMm}{R} - \dfrac{GMm}{3R}$

$V = \dfrac{2GMm}{3R}$
Thus work done to increase the radius of orbit of satellite is

$\dfrac{2GMm}{3R}$

Energy required to move a body of mass m from an orbit of radius 2R to 3R is:

0%
$\dfrac{GMm}{12R^{2}}$
0%
$\dfrac{GMm}{3R^{2}}$
0%
$\dfrac{GMm}{8R}$
0%
$\dfrac{GMm}{6R}$

Explanation

$\text{Energy required} =\text{(Final potential-Initial potential)m}$

$\text{Energy required}=-\dfrac{GMm}{3R}-\dfrac{-GMm}{2R}=\dfrac{GMm}{6R}$

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

Explanation

Value of g is given by:

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

Initial Potential Energy:

${ U }_{ i }=-\dfrac { { GM }_{ e }m }{ { R } } =-MgR$

Final Potential Energy:

$U_f =-\dfrac { { GM }_{ e }m }{ { 2R } } =-MgR/2$
Therefore Gain in Potential Energy:

${ U }_{ f }-{ U }_{ i }=-MgR/2-(-MgR)=\dfrac{MgR}{2}$

The gravitational P.E. of a rocket of mass 100 kg at a distance of

$10^{7}$ m from the earths centre is

$-4\times 10^{9}$ J. The weight of the rocket at a distance of

$10^{9}$ m from the centre of the earth is :

0%
$4\times 10^{-2}$ N
0%
$4\times 10^{-9}$ N
0%
$4\times 10^{-6}$ N
0%
$4\times 10^{-3}$ N

Explanation

Gravitational potential energy is given by

$PE=-\dfrac {GMm}{R}=-4\times 10^9$

But

$R = 10^7$ and

$PE=-4\times 10^9$ J

$\Rightarrow GMm=4\times10^9\times10^7=4\times10^{16}$
Now the weight of the satellite at a radius of

$10^9m$ from earth's centre is given by,

$mg=\dfrac {GMm}{R^2}=\dfrac {4\times10^{16}}{10^{18}}=4\times10^{-2}$ N

The angular velocity of rotation of a star (mass M and radius R), such that the matter will start escaping from its equator is:

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

Explanation

$v_{esc} = \sqrt{ \dfrac{2GM}{R}}$

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

$\therefore \omega = \sqrt{ \dfrac{2GM}{R^3}}$

$\omega =\displaystyle \dfrac{1}{R}\sqrt{\dfrac{2GM}{R}}$

A small body is initially at a distance

$r$ from the centre of earth.

$r$ is greater than the radius of the earth. If it takes

$W$ joules of work to move the body from this position to another position at a distance

$2r$ measured from the centre of earth, then how many joules would be required to move it from this position to a new position at a distance of

$3r$ from the centre of the earth.

0%
$\dfrac{W}{5}$
0%
$\dfrac{W}{3}$
0%
$\dfrac{W}{2}$
0%
$\dfrac{W}{6}$

Explanation

Work done to move the body in the gravitational field is equal to change in the potential energy of the body.
i.e.

$W=-\dfrac {GMm}{2r}-\left(-\dfrac {GMm}{r}\right)$

$\Rightarrow W = \dfrac {GMm}{2r}$
Now when the body is moved from

$2r$ to

$3r$ , change in PE is

$W'=-\dfrac {GMm}{3r}-\left(-\dfrac {GMm}{2r}\right)=\dfrac {1}{3}\dfrac {GMm}{2r}=\dfrac {W}{3}$

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

Explanation

Gravitational potential energy on the earth surface, $U_{r}=\displaystyle \dfrac{-GMm}{R}$

Gravitational potential energy at a height h above the earth's surface, $U_{h}=\displaystyle \dfrac{-GMm}{R+h}$

$U_{h}=\displaystyle \dfrac{-GMm}{R+R}=\dfrac{-GMm}{2R}$

Gain in gravitational potential energy $=U_{h}-U_{r}$

$=\displaystyle \dfrac{-GMm}{2R}-(\dfrac{-GMm}{R})=\dfrac{GMm}{R}-\dfrac{GMm}{2R}$

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

A person bring a mass of

$1 kg$ from infinite to point

$A$ . Initially the mass was at rest but is moves a speed of

$2 m /s$ as it reaches to

$A$ . The workdone by the person on mass is

$-3 J$ the gravitational potential at

$A$ is

0%
$-3 J / kg$
0%
$-2 J / kg$
0%
$-5 J / kg$
0%
$-7 J / kg$

Explanation

$K.E_{f}=1.2^2/2=2$
Work Energy Theorem,

$E_{f}-E_{i}=W$

$2+P.E-0=-3$

$P.E=mV=-5$

$V=-5$ since

$m=1$

The change in the P.E., when a body of mass

$m$ is displaced from Earth's surface to a vertical height equal to radius of earth (g

$=$ acceleration due to gravity on earth surface) is:

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

Explanation

$\displaystyle \Delta PE=\frac{GMm}{R}\left \{ 1-\frac{1}{2} \right \}= \displaystyle \frac{GMm}{2R}$

$gR=\displaystyle \frac{GM}{R}$

$\displaystyle \Delta PE=\frac{mgR}{2}$

What is the minimum energy required to launch a satellite of mass

$m$ from the surface of a planet of mass

$M$ and radius

$R$ in a circular orbit at an altitude of

$2R$ ?

0%
$\displaystyle \frac{2GmM}{3R}$
0%
$\displaystyle \frac{GmM}{2R}$
0%
$\displaystyle \frac{GmM}{3R}$
0%
$\displaystyle \frac{5GmM}{6R}$

Explanation

Given that,

Mass of satellite $=m$

Mass of planet $=M$

Radius $=R$

Altitude $h=2R$

Now,

The gravitational potential energy

$P.E=\dfrac{-Gm}{r}$

Potential energy at altitude $=\dfrac{GmM}{3R}$

Orbital velocity ${{v}_{0}}=\sqrt{\dfrac{2GmM}{R+h}}$

Now, the total energy is

${{E}_{f}}=\dfrac{1}{2}mv_{0}^{2}-\dfrac{GmM}{3R}$

${{E}_{f}}=\dfrac{1}{2}\dfrac{GmM}{3R}-\dfrac{GMm}{3R}$

${{E}_{f}}=\dfrac{GmM}{3R}\left[ \dfrac{1}{2}-1 \right]$

${{E}_{f}}=\dfrac{-GmM}{6R}$

Now, ${{E}_{i}}={{E}_{f}}$

Now, the minimum required energy

$K.E=\dfrac{Gmm}{R}-\dfrac{GmM}{6R}$

$K.E=\dfrac{5GmM}{6R}$

Hence, the minimum required energy is $\dfrac{5GmM}{6R}$

A sky laboratory of mass

$2\times 10^{3}Kg$ is raised from a circular orbit of radius 2R to a circular orbit of radius 3R. The work done is (approximately):

0%
$10^{16}J$
0%
$2\times 10^{10}J$
0%
$10^{6}J$
0%
$3\times 10^{10}J$

Explanation

$W=U_{2}-U_{1}=\displaystyle \frac{+Mm}{3R}(-\frac{GMm}{2R})=\frac{GMm}{6R}=\frac{mgR}{6}$

$= W=\displaystyle \frac{2\times 10^{3}\times 9.8\times 6.4\times 10^{6}}{6}=2\times 10^{10}J$

A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The magnitude of the gravitational potential at a point situated at a/2 distance from the centre will be :

0%
$\frac{GM}{a}$
0%
$\frac{2GM}{a}$
0%
$\frac{3GM}{a}$
0%
$\frac{4GM}{a}$

Explanation

gravitational potential at x point

$V_{x}=\frac{GM}{a/2}+\frac{GM}{a}=\frac{3GM}{a}$

A particle of mass M is situated at the center of a spherical shell of same mass and radius a. The gravitational potential at a point situated at

$\frac{a}{2}$ distance from the centre, will be

0%
$-\frac{3GM}{a}$
0%
$-\frac{2GM}{a}$
0%
$-\frac{GM}{a}$
0%
$-\frac{4GM}{a}$

The kinetic energy needed to project a body of mass

$m$ from the earth surface to infinity is

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

Explanation

Kinetic energy required

$=\displaystyle \frac{1}{2}mV_{e}^{2}$

$=\displaystyle \frac{1}{2}m(\sqrt{2gR})^{2} = mgR$

A satellite of mass

$m$ moves along an elliptical path around the earth. The areal velocity of the satellite is proportional to

0%
$m$
0%
$m^{-1}$
0%
$m^{0}$
0%
$m^{\frac{1}{2}}$

Explanation

The areal velocity

$\dfrac{dA}{dt}$

$=$ constant

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

$L=mvr \sin \theta$ (angular momentum)

$\dfrac{dA}{dt}=\dfrac{mvr \sin \theta}{2m}$

$= v r \sin \theta$

$\therefore \dfrac{dA}{dt} \ \alpha\ m^0$

$V_{e}$ is the escape velocity of a body from a planet of mass M and radius R. Then the velocity of the satellite revolving at height h from the surface of the planet will be:

0%
$V_{e}\sqrt{\dfrac{R}{R+h}}$
0%
$V_{e}\sqrt{\dfrac{2R}{R+h}}$
0%
$V_{e}\sqrt{\dfrac{R+h}{R}}$
0%
$V_{e}\sqrt{\dfrac{R}{2\left ( R+h \right )}}$

Explanation

Escape velocity,

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

and orbital velocity,

$v_o = \sqrt{\dfrac{GM_e}{(R+h)}}$

On dividing both equations,

$v_o = v_e \sqrt{\dfrac{R}{2(R+h)}}$

The height at which the acceleration due to gravity becomes

$\displaystyle \frac{\mathrm{g}}{9}$ (where

$\mathrm{g}=$ the acceleration due to gravity on the surface of the earth) in terms of

$\mathrm{R}$ , the radius of the earth, is:

0%
$2\ \mathrm{R}$
0%
$\displaystyle \frac{\mathrm{R}}{\sqrt{2}}$
0%
$\mathrm{R}/2$
0%
$\sqrt{2}\ \mathrm{R}$

Explanation

$\displaystyle \frac{\mathrm{G}\mathrm{M}}{9\mathrm{R}^{2}}=\frac{\mathrm{G}\mathrm{M}}{(\mathrm{R}+\mathrm{h})^{2}}$

$\Rightarrow 3\mathrm{R}=\mathrm{R}+\mathrm{h}$

$\Rightarrow \mathrm{h}=2\mathrm{R}$

The ratio of Earth's orbital angular momentum (about the sun) to its mass is

$4.4 \times 10^{15}\ m^{2}\ s^{-1}$ . The area enclosed by the earth's orbit is approximately

0%
$1 \times 10^{22}\ m^{2}$
0%
$3 \times 10^{22}\ m^{2}$
0%
$5 \times 10^{22}\ m^{2}$
0%
$7 \times 10^{22}\ m^{2}$

Explanation

Area enclosed by a body in an orbit is given by:

$\dfrac {dA}{dt}=\dfrac {L}{2m}$
Total orbital area enclosed by the earth is the area covered in one revolution i.e 365days.

$\therefore A=\dfrac {L}{2m}365\times 24\times 3600=\dfrac {4.4\times10^{15}}{2}365\times 86400=6.938\times 10^{22}$ m

$^2$

At what height, is the value of g half that on the surface of earth? (R = radius of the earth)

0%
0.414R
0%
R
0%
2R
0%
3.5R

Explanation

We know, g/g' =

$r^2$ /

$R^2$ , where g is gravitational force on earth, g' is gravitational force at a height.
Given g' = g/2

$g' = \dfrac{gR^2}{r^2} = \dfrac{g}{2}$

$\Rightarrow r^2 = 2R^2$

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

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

$= R(1.414-1)$

$= 0.414 R$

Experimental value of G is

0%
$6.67\times 10^{-11} Nm^2/kg^2$
0%
$66.7\times 10^{-11} Nm^2/kg^2$
0%
$667\times 10^{-11} Nm^2/kg^2$
0%
None of these

Maximum weight of a body is

0%
at the center of the earth.
0%
inside the earth.
0%
on the surface of the earth.
0%
above the surface of earth.

Explanation

$\textbf{Hint:}$ Gravitational acceleration vary with height or depth.

$\textbf{Step1:Weight of body}$

Weight of a body at any place is given by,

$W = mg$

Where,

$m$ is mass of the body,

g is the value of acceleration due to gravity at the place where body is kept.

$\textbf{Step2:Variation in gravitational acceleration}$

Let the value of acceleration due to gravity at surface of Earth is

$g$ .

$\bullet$ If radius of Earth is $R$ , the value of acceleration due to gravity at height $h$ from the surface of Earth is given by the formula,

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

$\bullet$ The value of acceleration due to gravity at depth

$h$ from surface of Earth is given by the formula,

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

$\bullet$ Therefore, the value of acceleration due to gravity is maximum at the surface of Earth.

Thus, the weight of the body $( W=mg )$ is maximum at the surface of Earth.

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

$\omega_1 , \omega_2, \omega_3$ then

0%
${\omega}_1 < {\omega}_2 > {\omega}_3$
0%
${\omega}_1 = {\omega}_2 = {\omega}_3$
0%
${\omega}_1 < {\omega}_2 < {\omega}_3$
0%
${\omega}_1 > {\omega}_2 > {\omega}_3$

Explanation

For position

$(1)$ :

$g_1=\dfrac{GM}{R^3}(R-d)$ ,

$w_1=mg_1$

For position

$(2)$ :

$g_2=\dfrac{GM}{R^2}$ ,

$w_2=mg_2$

For position

$(3)$ :

$g_3=\dfrac{GM}{(R+h)^2}$ ,

$w_3=mg_3$

Clearly we can see that

$g_1 < g_2 > g_3$

Thus

$w_1 < w_2 > w_3$ (A).

At a place, the value of 'g' is less by 1% than its value on the surface of the Earth (Radius of Earth,

$R=6400 km$ ). The place is :

0%
64 km below the surface of the earth
0%
64 km above the surface of the earth
0%
30 km above the surface of the earth
0%
32 km below the surface of the earth

Explanation

Let acceleration due to gravity at a place be

$g_d$ . If g is acceleration due to gravity at the surface then,

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

$0.99g=g(1-\frac {d}{R})$

$\frac {d}{R}=(0.01)$

$d=0.01\times 6400=64 km$ below the surface of earth

Depth from the surface of the earth at which is acceleration due to gravity is 25% of acceleration due to gravity at the surface

0%
1200 km
0%
4000 km
0%
3600 km
0%
4800 km

Explanation

g at the surface is given as

$g_s=\dfrac{4}{3}\pi GR\rho$
and at a depth d it is given as

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

$\dfrac{g_d}{g_s}=\dfrac{R-d}{R}$
Thus for the depth where acceleration is 25% of the surface gravity we get

$g_d$ as

$\dfrac{g}{4}$
or

$\dfrac{g}{4}=g\left ( 1-\dfrac{d}{R} \right )\Rightarrow d=\dfrac{3}{4}\times R = 4800 km$ (R is 6371 km)

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%
mg 2R
0%
$\displaystyle \frac{2}{3}mgR$
0%
3 mgR
0%
$\displaystyle \frac{1}{3}mgR$

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

0%
zero
0%
infinite
0%
$R$ times the weight at the surface of the earth
0%
$1/R^{2}$ times the weight at surface of the earth

A man weighs

$60 kg$ at earth's surface. At what height above the earth's surface his weight becomes

$30 kg$ ? (radius of earth

$=6400 km$ ).

0%
$1624 km$
0%
$2424 km$
0%
$2624 km$
0%
$2826 km$

Explanation

Above the earth's surface,

$g_1 = \dfrac{g}{(1+\dfrac{h}{R})^2}$
But changed weight = half of its weight on the earth.

$\Rightarrow g_1 = \dfrac{g}{2}$

$\therefore h = 2624 km.$

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

$g_h$ at height h is given by

$g_h = \dfrac{GM}{(R+h)^2} = GM(R+h)^{-2}$ which denotes a parabola

$g_x$ at depth x is given by

$g_x = g \left [ \dfrac{R-x}{R} \right ]$
Which denotes is straight line. So, option A is correct.

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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