MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 14

A body is projected upwards with a velocity of

$4\times 11.2 km/s$ from the surface of earth. What will be the velocity of the body when it escapes from the gravitational pull of earth?

0%
$11.2\ km/s$
0%
$2\times 11.2\ km/s$
0%
$3\times 11.2\ km/s$
0%
$\sqrt {15}\times 11.2\ km/s$

Suppose, the acceleration due to gravity at the Earth's surface is

$10m s^{-2}$ and at the surface of Mars it is

$4.0 m s^{-2}$ . A

$60$ kg passenger goes from the Earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of figure best represents the weight (net gravitational force) of the passenger as a function of time.?

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

Explanation

$g\propto \displaystyle\frac{1}{R^2}$ so we will not get a straight line.
Also,

$F=0$ at a point where Force due to Earth=Force due to Mars.

A space vehicle approaching a planet has a speed v, when it is very far from the planet. At that moment tangent of its trajectory would miss the centre of the planet by distance R. If the planet has mass M and radius r, what is the smallest value of R in order that the resulting orbit of the space vehicle will just miss the surface of the planet?

0%
$\dfrac{r}{v}\left[v^2+\dfrac{2GM}{r}\right]^{\dfrac{1}{2}}$
0%
$vr\left[1+\dfrac{2GM}{r}\right]$
0%
$\dfrac{r}{v}\left[v^2+\dfrac{2GM}{r}\right]$
0%
$\dfrac{2GMv}{r}$

A planet revolves around the sun in an elliptical orbit of eccentricity e. If T is the time period of the planet, then the time spent by the planet between the ends of the minor axis and major axis close to the sun is

0%
$\dfrac{T\pi}{2e}$
0%
$T\left(\dfrac{2e}{\pi}-1\right)$
0%
$\dfrac{Te}{2\pi}$
0%
$T\left(\dfrac{1}{4}-\dfrac{e}{2\pi}\right)$

Explanation

As areal velocity of a planet around the sun is constant. Therefore, the desired time is

${t}_{AB}=\left(\dfrac{area ABS}{area \ of \ ellipse}\right)\times time \ period$

If a = semi-major axis and b= semi-minor axis of ellipse then, area of ellipse =

$\pi$ ab

Area

$ABS=\dfrac{1}{4}\left(area \ of \ ellipse \right)-Area \ of \ triangle ASO$

$=\dfrac{1}{4}\times\pi{ab}-\dfrac{1}{2}\left(ea\right)\times\left(b\right)$

$\therefore{t}_{AB}=\dfrac{\left[\dfrac{\pi\left(ab\right)}{4}-\dfrac{1}{2}eab\right]}{\pi ab}\times T=T\left(\dfrac{1}{4}-\dfrac{e}{2\pi}\right)$

1033810_936520_ans_5529b0a6d499487db20225c1ecdaa1ba.png

A projectile is fired vertically upwards from the surface of earth with a velocity of

$kv_{e}$ where

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

$K<1$ . Neglecting air resistance, the maximum height to which it will rise, measured from the centre of the earth, is (

$R_{E}$ -radius earth)

0%
$\dfrac{R_{E}}{1-k^{2}}$
0%
$\dfrac{R_{E}}{k^{2}}$
0%
$\dfrac{1-k^{2}}{R_{E}}$
0%
$\dfrac{k^{2}}{R_{E}}$

Explanation

Let a projectile is fired vertically upwards from the earth's surface with a velocity v and reaches a height h. Energy of the projectile at the surface of the earth is

$E_{i}=\dfrac{1}{2}mv^{2}-\dfrac{GM_{E}m}{R_{E}}$

Energy of the projectile at a height h

$E_{f}=-\dfrac{GM_{E}m}{R_{E}+h}$ (

$\therefore$ At height h velocity of the projectile is zero.)

According to law of conservation of energy,

$E_{i}=E_{f}$

$\dfrac{1}{2}mv^{2}-\dfrac{GM_{E}m}{R_{E}}=-\dfrac{GM_{E}m}{R_{E}+h}$

$\dfrac{1}{2}mv^{2}=GM_{E}m\left[\dfrac{1}{R_{E}}-\dfrac{1}{R_{E}+h}\right]=\dfrac{GM_{E}mh}{\left(R_{E}\right)\left(R_{E}+h\right)}$

$\dfrac{1}{2}mv^{2}=\dfrac{mghR_{E}}{R_{E}+h} \left(\because g=\dfrac{GM_{E}}{R_{E}^{2}}\right)$

As per question,

$v=kv_{e}=k\sqrt{2gR_{E}} \left(\because v_{e}=\sqrt{2gR_{E}}\right)$

and

$h=r-R_{E}$

$\therefore\dfrac{1}{2}mk^{2}2gR_{E}=\dfrac{mg\left(r-R_{E}\right)R_{E}}{r}$

$k^{2}=\dfrac{r-R_{E}}{r}\Rightarrow\left(1-k^{2}\right)r=R_{E}$

$\Rightarrow r=\dfrac{R_{E}}{1-k^{2}}$

If g is the acceleration due to gravity on the surface of earth, find the gain in potential energy of an object of mass m raised from the surface of earth to a height equal to the radius R of the earth.

0%
$\dfrac{1}{2}mg R$ .
0%
$\dfrac{1}{4}mg R$ .
0%
$\dfrac{1}{3}mg R$ .
0%
$\dfrac{2}{1}mg R$ .

Explanation

Let M, R be the mass and radius of the earth. Then

$g=GM/R^2$
or

$GM=gR^2$ (i)
Potential energy of the object on the surface of earth

$U_1=\dfrac{-GMm}{R}$
The potential energy of the object at a height equal to radius of the earth is

$U_2=\dfrac{-GMm}{2R}$
Gain in PE is

$U_2-U_1=-\dfrac{GMm}{2R}+\dfrac{GMm}{R}=\dfrac{GMm}{2R}$

$=\dfrac{(gR^2)m}{2R}=\dfrac{1}{2}mg R$ .

A particle of mass m (starting from rest) moves vertically upwards from the surface of earth under an external force

$\underset{F}{\rightarrow}$ which varies with height z as,

$\underset{F}{\rightarrow}=(2-\alpha z)m \underset{g}{\rightarrow}$ , where

$\alpha$ is a positive constant. If H is the maximum height to which particle rises, then which of the following is / are correct ?

0%
$H=\dfrac{2}{\alpha}$
0%
Work done by $\underset{F}{\rightarrow}$ during motion upto $\dfrac{H}{2}$ is $\sqrt{\dfrac{3mg}{2 \alpha}}$ .
0%
Velocity of particle at $\dfrac{H}{2}$ is $\sqrt{\dfrac{g}{\alpha}}$
0%
$=3\sqrt{\cfrac{GM_e}{5R}}$

Explanation

A particle of mass

$=m$ moves vertically upwards from the surface of earth under an external force

$\overrightarrow{F}$

Let, m be the mass of the body

$M_e$ be the mass of the earth.

$R_e$ be the radius of the earth.

According to the question,

Let,

$U$ be the gravitation potential energy a height which is equal to times of the radius of earth.

$U=-\cfrac{GM_em}{R+qR}=-\cfrac{GM_em}{10R}$

Increase in

$U=-\cfrac{GM_em}{10R}\\=-\cfrac{GM_em}{R}\\ -\cfrac{GM_em}{R}(-\cfrac{1}{10}+1)\\=-\cfrac{9GM_em}{10R}$

Increase in potential energy

$=$ Decrease in finite energy

$\cfrac{9GM_em}{10R}=\cfrac{1}{2}mv^2\\v^2=\cfrac{9\times GM_e}{10R}=\sqrt{\cfrac{9GM_e}{5R}}\\v=3\sqrt{\cfrac{GM_e}{5R}}$

Hence projected velocity will

$=3\sqrt{\cfrac{GM_e}{5R}}$

A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of radius

$2$ R(from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius

$3$ R. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are.

0%
$\dfrac{3}{4}mgR, \dfrac{mgR}{6}$
0%
$\dfrac{3}{4}mgR, \dfrac{mgR}{12}$
0%
$mgR, mgR$
0%
$2mgR, mgR$

Explanation

Energy required to place the lab in first orbit is the change in potential energy

$+{ k }_{ 0 }{ E }_{ 0 }$

$\triangle PE=\cfrac { -GMm }{ 2R } +\cfrac { GMm }{ 2R } =\cfrac { GMm }{ 2R } =\triangle u$

Velocity given to lab is provided by gravitation force. So, centripetal force=gravitational force

$\cfrac { m{ v }^{ 2 } }{ 2R } =\cfrac { GMm }{ { \left( 2R \right) }^{ 2 } } \Rightarrow v=\sqrt { \cfrac { GM }{ 2R } }$

$KE=\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { 1 }{ 2 } \cfrac { GM }{ 2R } =\cfrac { 1 }{ 4 } \cfrac { GMm }{ R }$

Total energy

$=\triangle u+{ K.E }_{ 0 }$

$=\cfrac { GMm }{ 2R } +\cfrac { 1 }{ 4 } \cfrac { GMm }{ R }$

$=\cfrac { 3 }{ 4 } \cfrac { GMm }{ R }$

$g=\cfrac { GMm }{ { R }^{ 2 } }$

$T.E.=\cfrac { 3 }{ 4 } mgR$

Similarly to shift to orbit of

$3R$

$\cfrac { m{ v }^{ 2 } }{ 3R } =\cfrac { GMm }{ { \left( 3R \right) }^{ 2 } } \Rightarrow { v }^{ 2 }=\cfrac { GM }{ 3R }$

Total energy

$=\cfrac { GMm }{ R } -\cfrac { GMm }{ 3R } +\cfrac { 1 }{ 2 } m{ v }^{ 2 }$

$=\cfrac { GMm }{ R } -\cfrac { GMm }{ 3R } +\cfrac { 1 }{ 2 } \cfrac { GMm }{ 3R }$

$=\cfrac { GMm }{ R } \left( 1-\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 } \right)$

$=\cfrac { 5 }{ 6 } mgR$

Energy required

$=\cfrac { 5 }{ 6 } mgR-\cfrac { 3 }{ 4 } mgR$

$=\cfrac { \left( 20-18 \right) mgR }{ 24 }$

Minimum energy required

$=\cfrac { mgR }{ 12 }$

A point mass m is released from rest at a distance of

$3$ R from the centre of a thin-walled hollow sphere of radius R and mass M as shown. The hollow sphere is fixed in position and the only force on the point mass is the gravitational attraction of the hollow sphere. There is a very small hole in the hollow sphere through which the point mass falls as shown. The velocity of a point mass when it passes through point P at a distance

$R/2$ from the centre of the sphere is?

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$\sqrt{\dfrac{2GM}{3R}}$
0%
$\sqrt{\dfrac{5GM}{3R}}$
0%
$\sqrt{\dfrac{25GM}{24R}}$
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None of these

Explanation

Gravitational potential inside hollow sphere

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

Gravitational potential at distance

$3R=-\cfrac{GM}{3R}$

$\therefore$ Conservation of energy

$\Rightarrow\cfrac{1}{2}mv^2-\cfrac{GMm}{R}=-\cfrac{GMm}{3R}\\ \Rightarrow {1}{2}mv^2=\cfrac{GMm}{R}(1-\cfrac{1}{3})\\ \Rightarrow\cfrac{v^2}{2}=\cfrac{2GM}{3R}\\ \Rightarrow v^2=\cfrac{4GM}{3R}\\ v=\sqrt{\cfrac{4GM}{3R}}$

1117853_986672_ans_2f0593d7a8004fcfa1a4424481c4de21.png

What is the fractional decrease in the value of free-fall acceleration g for a particle when it is lifted from the surface to an elevation h? (h

$< <$ R).

0%
$-2\left(\dfrac{h}{R}\right)$ .
0%
$-3\left(\dfrac{h}{R}\right)$ .
0%
$-4\left(\dfrac{h}{R}\right)$ .
0%
$-5\left(\dfrac{h}{R}\right)$ .

Explanation

Given that

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

$\dfrac { dg }{ dR } =\dfrac { -2GM }{ { R }^{ 3 } }$

$\dfrac { dg }{ h } =\dfrac { -2GM }{ { R }^{ 2 } } .\dfrac { 1 }{ R }$

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

KEPLER'S LAWS
A satellite is in an elliptic orbit around the earth with aphelion of

${6R}_{E}$ and perihelion of

${2R}_{D}$ where

${R}_{E}$ is the radius of the earth. The eccentricity of the orbit is:

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$

Explanation

Here,

${r}_{A}={6R}_{E}, {r}_{P}={2R}_{E}$
The eccentricity of the orbit is

$e=\dfrac{{r}_{A}-{r}_{P}}{{r}_{A}+{r}_{P}};$

$e=\dfrac{{6R}_{E}-{2R}_{E}}{{6R}_{E}+{2R}_{E}}$

$=\dfrac{4}{8}=\dfrac{1}{2}$

1014203_936526_ans_2ec2ce311bbb483ea65715d28bfbb7b8.png

In a hypothetical uniform and spherical planet of mass

$M$ and radius

$R$ , a tunnel is dug radially from its surface to its centre as shown. The minimum energy required to carry a unit mass from its centre to the surface is

$kmgR$ . Find value of

$k$ . Acceleration due to gravity at the surface of the planet is

$g$ .

0%
0.7
0%
0.5
0%
1
0%
0.9

If a mass

$m$ is placed in the vicinity of another mass

$M$ , It experiences a gravitational force of attraction. However, if these two masses are at a very large distance, the force of attraction is negligible. The gravitation potential is the amount of work done per unit mass in bringing it slowly from infinity to some finite distance from

$M$ . Hence, gravitational potential is a state function rather than a path function.
One application of this potential is in finding the escape velocity of a body from earth. As we know the gravitation potential energy associated with mass

$m$ on earth surface is

$-\dfrac {GMm}{R}$ . The mechanical energy conservation gives

$\underline {V_{escape} = \sqrt {\dfrac {2GM}{R}}}$ or

$V_{escape} = \sqrt {2gR}$ . An inquisitive mind decides to get the result by kinematics, considering a particle following curvilinear path from surface to infinity, gravity changes with height as

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

$v\dfrac {dv}{dh} = -g\left (1 + \dfrac {h}{R}\right )^{-2}$ . Solution of the equation within limits

$h = 0$ to

$h = \infty$ gives

$\underline {V_{escape} = \sqrt {2gR}}$ .
For most of the objects, either point masses or objects of finite dimension, the variation of potential in space exhibits symmetric behaviour. In case of spherical objects of uniform density, the locus of equipotential points is a spherical shell of any given radius. Hence, this potential field is symmetric about all three axes passing through the centre of spherical object. Consider a somewhat complicated objects as shown in the adjacent figure.
The figure shows a solid cube of edge length

$10\ cm$ . The origin is the centre of cube as shown. Eight spherical cavities are formed in this cube, each having a radius of

$1\ cm$ and centers at

$(\pm 2\ cm, \pm 2\ cm,\pm 2\ cm$ ). This figure shows wide range of equipotential surfaces / curves.
If a particle is thrown with escape velocity

$(v_{e})$ from surface of earth, which of the following is the correct energy equation?

0%
$\dfrac {1}{2}mv_{e}^{2} - \dfrac {GMm}{2R} = -\dfrac {GMm}{R}$
0%
$\dfrac {1}{2}mv_{e}^{2} - \dfrac {GMm}{2R} = 0$
0%
$\dfrac {1}{2}mv_{e}^{2} - \dfrac {GMm}{R} = 0$
0%
None of these

The magnitudes of the gravitational field at distances

$r_1$ and

$r_2$ from the centre of a uniform sphere of radius

$R$ and mass

$M$ are

$F_1$ and

$F_2$ , respectively. Then.

0%
$\dfrac{F_1}{F_2}=\dfrac{r_1}{r_2}$ if $r_1 < R$ and $r_2 < R$
0%
$\dfrac{r^2_2}{r_2}$ if $r_1 > R$ and $r_2 > R$
0%
$\dfrac{F_1}{F_2}=\dfrac{r_1}{r_2}$ if $r_1 > R$ and $r_2 > R$
0%
$\dfrac{F_1}{F_2}=\dfrac{r^2_1}{r^2_2}$ if $r_1 < R$ and $r_2 < R$

Explanation

For a solid sphere of mass M and radius R -

$\Rightarrow$ Gravitational field

$=\dfrac{GM}{r^2}$ (

$r>$ Radius of sphere )

$=\dfrac{GMr}{R^3}$

$\left( r<R\right)$

$*$ If

$r_1<R$ and

$r_2<R$

$\Rightarrow F_1=\dfrac{GMr_1}{R^3}$ and

$F_2=\dfrac{GMr_2}{R^3}$

$\Rightarrow \dfrac{F_1}{F_2}=\dfrac{r_1}{r_2}$

Option (A) is correct.

$*$ If

$r_1<R$ and

$r_2<R$

$\Rightarrow F_1=\dfrac{GM}{R_1^2}$ and

$F_2=\dfrac{GM}{R^2_2}$

$\Rightarrow \dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}$

Hence, option A and B correct.

The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R and the radius of the planet is R'. Then what is the value of R/R'?

0%
2
0%
3
0%
4
0%
5

Explanation

let,

$\\ { g }^{ \prime }$ = aceleration due to gravity at newly discovered planet.

g=acceleration due to gravity at earth

$\\ \therefore { g }^{ \prime }=g\\ =>\cfrac { G{ M }^{ \prime } }{ { R }^{ \prime 2 } } =\cfrac { GM }{ { R }^{ 2 } } \\ =>\cfrac { G.\cfrac { 4 }{ 3 } \pi { R }^{ \prime 3 }{ p }^{ 1 } }{ { R }^{ \prime 2 } } =\cfrac { G.\cfrac { 4 }{ 3 } \pi { R }^{ 2 }{ p } }{ { R }^{ 2 } } \\ =>{ R }^{ \prime }{ p }^{ \prime }=R.p\\ =>\cfrac { R }{ { R }^{ \prime } } =\cfrac { { p }^{ \prime } }{ p } =\cfrac { 2p }{ 2 } =2$

Use the assumptions of the previous question. An object weighed by a spring balance at the equator gives the same reading as a reading taken at a depth

$d$ below the earth's surface at a pole

$(d < < R)$ . The value of

$d$ is

0%
$\dfrac {\omega^{2}R^{2}}{g}$
0%
$\dfrac {\omega^{2}R^{2}}{2g}$
0%
$\dfrac {2\omega^{2}R^{2}}{g}$
0%
$\dfrac {\sqrt {Rg}}{\omega}$

Imagine a new planet having the same density as that of the earth but it is

$3$ times bigger than the earth is size. If the acceleration due to gravity on the surface of the earth is g and that on the new planet is g', then what is the value of g'/g?

0%
3
0%
4
0%
5
0%
6

Explanation

let

${ g }^{ \prime }$ and

$g$ is the acceleration due to gravity at the new planet and at Earth.

$\cfrac { { g }^{ \prime } }{ g } =\cfrac { \cfrac { G{ M }^{ \prime } }{ { R }^{ \prime 2 } } }{ \cfrac { GM }{ { R }^{ 2 } } } =\left( \cfrac { { M }^{ \prime } }{ M } \right) \times \left( \cfrac { R }{ { R }^{ \prime } } \right) ^{ 2 }=\left( \cfrac { { V }^{ \prime }P }{ VP } \right) \left( \cfrac { R }{ { R }^{ \prime } } \right) ^{ 2 }\quad \left[ { V }^{ \prime }=3V\\ =>p.\cfrac { 4 }{ 3 } \pi { R }^{ \prime 3 }=3p.\cfrac { 4 }{ 3 } \pi { R }^{ 3 }\\ =>{ R }^{ \prime 3 }=3{ R }^{ 3 }\\ =>\cfrac { { R }^{ \prime } }{ R } =(3{ ) }^{ \cfrac { 1 }{ 3 } } \right] \\ =\cfrac { \cfrac { 4 }{ 3 } \pi { R }^{ \prime 3 } }{ \cfrac { 4 }{ 3 } \pi { R }^{ 3 } } \times \cfrac { { R }^{ 2 } }{ { R }^{ \prime 2 } } \\ =\cfrac { { R }^{ \prime } }{ R } \\ =(3{ ) }^{ \cfrac { 1 }{ 3 } }$

$g$ on the surface of the earth is

$9.8/ m{s}^{-2}$ , its value at a depth of

$3200\ km$ (Radius of the earth=

$6400\ km$ ) is

0%
$9.8/ m{s}^{-2}$ ,
0%
Zero
0%
$4.9/ m{s}^{-2}$ ,
0%
$2.45/ m{s}^{-2}$ ,

Explanation

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

$\implies 9.8\left(1-\dfrac{3200\times 10^{3}}{6400\times 10^{3}}\right)$

$4.9m/s^2$

A person on the surface of the moon:

0%
Does not feel the effect of earth's gravity because the gravity due to moon is such stronger.
0%
Does not feel the effect of earth's gravity.
0%
Does not feel the effect of earth's gravity and moon gravity he is freely towards the earth.
0%
Feels only the combined effect of earth's and moon gravity which are comparable in magnet.

Explanation

The value of ${g_e}$ on the surface of moon will be,

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

After putting the values of $g,h,R$

we get the value of ${g'}$ as,

${g'} = .0026\,{\rm{m}}/{{\rm{s}}^2}$

This is almost negligible and can’t be felt on the person standing on moon.

Thus, a person on the surface of the moon does not feel the effect of earth's gravity because the gravity due to moon is much stronger.

The depth at which the value of

$g$ becomes

$25$ % of that at the surface of the earth is (in Km)

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

Explanation

Given that

$g_d=0.25g$

We have,

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

$$\therefore 0.25g=g\left(1-\dfrac d{R}\right)$$

$\implies d=0.75R$

$\implies 0.75\times 6400$

$\implies 4800km$

For escape of oxygen molecules from the earth's surface, its temperature should be:

0%
$4.11\times {10}^{3}K$
0%
$2.09\times {10}^{4}K$
0%
$8.3\times {10}^{4}K$
0%
$3\times {10}^{5}K$

Explanation

rms velocity of gas molecule is given by,

$V_{rms}=\sqrt{\dfrac{3kT}{M}}=11.2\times 10^3$

$\therefore T=\dfrac{(11.2\times 10^3)^2\times}{3k}$

$\implies \dfrac{(11.2\times 10^3)^2(2.76\times 10^{-26})}{3\times 1.38\times 10^{-23}}$

$\implies 8.3\times 10^{4}k$

A particle hanging from a massless spring stretches it by

$2cm$ at the eath's surface. How much will the same particle stretch the spring at a height of

$2624Km$ from the surface of the earth? (Radius of the earth

$=6400Km$ )

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

Explanation

$\triangle l\alpha \quad F\\ \Rightarrow \triangle l\alpha \quad mg\\ \Rightarrow \triangle l\alpha \quad g\\ \therefore \dfrac { \triangle { l }_{ 2 } }{ \triangle { l }_{ 1 } } =\dfrac { { g }_{ 2 } }{ { g }_{ 1 } } \\ \Rightarrow \dfrac { \triangle { l }_{ 2 } }{ 2 } =\dfrac { \dfrac { GM }{ { \left( 6400+2624 \right) }^{ 2 } } }{ \dfrac { GM }{ { \left( 6400 \right) }^{ 2 } } } \\ \Rightarrow \triangle { l }_{ 2 }=1$

$\therefore \triangle { l }_{ 2 }$ will be

$1cm$ .

A rocket is launched normal to the surface of the earth, away from the sun, along the line joining the sun and the earth. The sun is $3 \times {10^5}$ times heavier than the earth and is at a distance $2.5 \times {10^4}$ times larger than the radius of the earth. The escape velocity from earths gravitational field is ${{\rm{V}}_{\rm{c}}}\;{\rm{ = 11}}{\rm{.2km}}{{\rm{s}}^{{\rm{ - 1}}}}$ . The minimum initial $\left( {{{\rm{V}}_{\rm{c}}}} \right)\;$ required for the rocket to be able to leave the sun-earth system is closest to (Ignore the rotation and revolution of the earth and the presence of any other planet)

0%
$\left( {{{\rm{V}}_{\rm{c}}}} \right)\;{\rm{ = 22km}}{{\rm{s}}^{{\rm{ - 1}}}}$
0%
$\left( {{{\rm{V}}_{\rm{c}}}} \right)\;{\rm{ = 42}}\;{\rm{km}}{{\rm{s}}^{{\rm{ - 1}}}}$
0%
$\left( {{{\rm{V}}_{\rm{c}}}} \right)\;{\rm{ = 62}}\;{\rm{km}}{{\rm{s}}^{{\rm{ - 1}}}}$
0%
$\left( {{{\rm{V}}_{\rm{c}}}} \right)\;{\rm{ = 72}}\;{\rm{km}}{{\rm{s}}^{{\rm{ - 1}}}}$

Explanation

$\dfrac { 1 }{ 2 } { mV }_{ e }^{ 2 }-\dfrac { { GM }_{ e }m }{ { R }_{ e } } -\dfrac { { GM }_{ e }m\times 3\times { 10 }^{ 5 } }{ 2.5\times { 10 }^{ 4 }{ R }_{ e } } =0$

$\dfrac { { Ve }^{ 2 } }{ 2 } =\dfrac { { GM }_{ m } }{ { R }_{ e } } \left[ \dfrac { 1+3\times { 10 }^{ 5 } }{ 2.5\times { 10 }^{ 4 } } \right]$

${ V }_{ e }=\sqrt { \dfrac { 13\left( { 2GM }_{ e } \right) }{ { R }_{ e } } } =\sqrt { 13 } \times 11.2\approx 42$

The minimum initial

${ V }_{ s }=42km/s$ .

The period of rotaion of the earth so as to make any object weightless on its equator is

0%
$84$ min
0%
$74$ minutes
0%
$64$ minutes
0%
$54$ minutes

Assuming the earth as a sphere of uniform density, the acceleration due to gravity half way towards the centre of the earth will be if it weighed 250N on the surface?

0%
$0.25\ g$
0%
$0.50\ g$
0%
$0.125\ g$
0%
$0.75\ g$

Explanation

Suppose

${ g }_{ 1 }{ g }_{ a }$ be the acceleration due to gravity on earth's surface and at a depth

$d$ from surface respectively. Also suppose

$w$ and

${ w }_{ d }$ be weight of body on earth's surface at a depth

$d$ .

$w=mg=250N\quad \longrightarrow \left( 1 \right)$

${ w }_{ d }={ mg }_{ d }\quad \longrightarrow \left( 2 \right)$

We know that

${ g }_{ d }=g\left( 1-\dfrac { d }{ R } \right) \quad \longrightarrow \left( 3 \right)$

Here

${ d }_{ 2 }=\dfrac { R }{ 2 }$

$R=$ radius of earth

From equation

$(3)$ and

$(4)$

${ g }_{ d }=g\left( 1-\dfrac { R/2 }{ R } \right)$

$=g\left( 1-1/2 \right) =g\times \dfrac { 1 }{ 2 } =\dfrac { g }{ 2 } \quad \longrightarrow \left( 5 \right)$

${ w }_{ d }={ mg }_{ d }=\dfrac { mg }{ 2 } =\dfrac { 1 }{ 2 } mg=\dfrac { 1 }{ 2 } w=\dfrac { 1 }{ 2 } \times 250=125N$ .

The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is

$16.0\ N$ when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.

0%
19.6 J
0%
29 J
0%
35 J
0%
10 J

Explanation

Given that,

Tension T = 16.0 N

Now,

$T-2mg+2ma=0....(I)$

$T-mg-ma=0....(II)$

From equation (I) and (II)

$3ma-mg=0$

$g=3a$

$a=\dfrac{g}{3}$

Now, put the value of g in equation (II)

$T-3ma-ma=0$

$T=4ma$

$a=\dfrac{T}{4m}$

Now, from equation of motion

At $t= 1$

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

$s=0+\dfrac{1}{2}\times \dfrac{T}{4m}\times {{(1)}^{2}}$

$s=\dfrac{16}{8m}$

$s=\dfrac{2}{m}$

Thus, change in the height of the block will be;

$\Delta h=s$

Net mass

$2m-m=m$

Now, decrease potential energy

$P.E=mg\Delta h$

$P.E=m\times 9.8\times \dfrac{2}{m}$

$P.E=19.6\,J$

Hence, the decrease potential energy is $19.6\ J$

If the mass and radius of a planet are doubled, then acceleration due to gravity on its surface will become :

0%
One fourth
0%
One half
0%
Double
0%
Four times

Explanation

The acceleration due to gravity on the surface is given as

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

If the mass and the radius are doubled then the acceleration due to gravity will become

$\dfrac{{G2m}}{{4{r^2}}}$

Hence the acceleration becomes one half.

$g_e$ and

$g_p$ denote the acceleration due to gravity on the surface of earth and another planet whose mass and radius are twice that of the earth, then

0%
$g_p = g_e$
0%
$g_p = \dfrac{g_e }{ 2}$
0%
$g_p = 2g_e$
0%
$g_p = 3g_e$

A body is released from a point distance r from the center of earth. If R is the earth r > R, then the velocity of the body at the time of striking the earth will be:

0%
$\sqrt{gR}$
0%
$\sqrt{2gR}$
0%
$\sqrt{\frac{2gR}{r - R}}$
0%
$\sqrt{\dfrac{2gR(r - R)}{r}}$

Explanation

${ \left( PE \right) }_{ height }={ \left( PE \right) }_{ surface }+{ \left( KE \right) }_{ surface }\\ -\dfrac { GMm }{ r } =-\dfrac { GMm }{ R } +\dfrac { 1 }{ 2 } m{ v }^{ 2 }\\ \therefore \dfrac { 1 }{ 2 } { v }^{ 2 }=GM\left( \dfrac { 1 }{ R } -\dfrac { 1 }{ r } \right) \\ \Rightarrow v=\sqrt { \dfrac { 2GM }{ R } \left( \dfrac { r-R }{ r } \right) } \dfrac { R }{ R } \\ \Rightarrow \sqrt { \dfrac { 2GMR }{ R } \left( \dfrac { r-R }{ r } \right) } \\ \Rightarrow \sqrt { \dfrac { 2gR(r-R) }{ r } }$

At what height the acceleration due to gravity will be reduced to

$36\%$ of its value on the surface of the earth?

0%
Use $g_{h} = g\left [\dfrac {2R}{R + h}\right ]^{2}$ .
0%
Use $g_{h} = g\left [\dfrac 3{R}{R + h}\right ]^{2}$ .
0%
Use $g_{h} = g\left [\dfrac {2R}{2R + h}\right ]^{2}$ .
0%
Use $g_{h} = g\left [\dfrac {R}{R + h}\right ]^{2}$ .

A body is projected with a velocity double the escape velocity. The velocity of the body after it escape the gravitational field of the earth is:

0%
$0$
0%
$V_{e}$
0%
$\sqrt{2} V_{e}$
0%
$\sqrt{3} V_{e}$

The decrease in the value of g at height h from earth's surface is

0%
$\dfrac{2h}{R}g$
0%
$\dfrac{\sqrt2h}{R}g$
0%
$\dfrac{h}{R}g$
0%
$\dfrac{R}{2hg}$

The Moon to orbits Jupiter once in

$1.769$ days. The orbital radius of the Moon to is

$421700$ km. Calculate the mass of Jupiter?

0%
$3.898\times { 10 }^{ 27 }kg$
0%
$1.898\times { 10 }^{ 27 }kg$
0%
$8.898\times { 10 }^{ 27 }kg$
0%
$5.898\times { 10 }^{ 27 }kg$

If the radius of earth's orbit is made

$1/4th$ , the duration of an year will become

0%
$8\ times$
0%
$4\ times$
0%
$1/8\ times$
0%
$1/4\ times$

Explanation

To solve this problem we have to resort to Kepler's Third law of orbital motion and it states that the square if average orbital period of a planet is proportional to the cube of the average orbital radius such that

${ T }^{ 2 }$

$\alpha$

${ R }^{ 3 }$

If we reduce radius of earth to form fold.

$\dfrac { { T }_{ 1 }^{ 2 } }{ { R }_{ 1 }^{ 3 } } =\dfrac { { T }_{ 2 }^{ 2 } }{ { \left( \dfrac { { R }_{ 1 } }{ 4 } \right) }^{ 3 } }$

$\dfrac { { T }_{ 1 }^{ 2 } }{ { R }_{ 1 }^{ 3 } } =\dfrac { { 64T }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 3 } }$

${ T }_{ 2 }^{ 2 }=\dfrac { { T }_{ 1 }^{ 2 } }{ 64 }$

Taking square root of both sides

$\sqrt { { T }_{ 2 }^{ 2 } } =\sqrt { \dfrac { { T }_{ 1 }^{ 2 } }{ 64 } }$

${ T }_{ 2 }=\dfrac { { T }_{ 1 } }{ 8 }$

Orbital period of earth duration of gear

$\dfrac { 1 }{ 8 }$ times.

A man weighs

$100\ kg$ on the surface of the earth of radius

$R$ . At what height above the surface of the earth, he will weigh

$50\ kg$ ?

0%
$0.41\ R$
0%
$0.51\ R$
0%
$0.31\ R$
0%
$0.61\ R$

A narrow tunnel is dug across a planet diametrically and small bod is dropped from a large height, so that it falls the tunnel. The variation of its kinetic energy 'E' with distance r, from the centre is represented by:

A satellite of mass

$'m'$ is revolving around the earth in an orbit of radius

$10R$ (

$R$ is the radius of earth), the minimum energy required to give the satellite so that it can escape the gravitational field of the earth is [mass of earth

$=M_e$ ]:

0%
$\dfrac{{G{M_e}m}}{{20R}}$
0%
$\dfrac{{G{M_e}m}}{{10R}}$
0%
$\dfrac{{G{M_e}m}}{{R}}$
0%
$\dfrac{{G{M_e}m}}{{2R}}$

At what depth below the surface of the earth, is the value of

$g$ same as that at a height of

$10\ km$ from the surface of the earth?

0%
$5\ km$
0%
$10\ km$
0%
$20\ km$
0%
$40\ km$

If the gravitational acceleration at the Earth's surface is

$9.81\,m{s^{ - 2}}$ , what is its value at a height equal to the diameter of the Earth from its surface?

0%
$4.905\,m{s^{ - 2}}$
0%
$2.42\,m{s^{ - 2}}$
0%
$3.27\,m{s^{ - 2}}$
0%
$1.09\,m{s^{ - 2}}$

Choose the correct answer:
The height over the surface of the earth at which the gravitational field of the earth becomes

$\frac{1}{4}$ th of the field at the surface

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

The kinetic energy needed to project a body of mass m from the earth's surface (radius

$R$ ) to infinity is -

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

A rubber ball is dropped from a height of

$5$ cm on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to

$1.8$ m. The ball loses its velocity on bouncing by a factor of?

0%
$\dfrac{16}{25}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{9}{25}$

Explanation

$\dfrac { \dfrac { 1 }{ 2 } { { mv }_{ 1 } }^{ 2 } }{ \dfrac { 1 }{ 2 } { { mv }_{ 2 } }^{ 2 } } =\dfrac { mg{ h }_{ 1 } }{ mg{ h }_{ 2 } } =\dfrac { 25 }{ 9 }$

$\dfrac { { { v }_{ 1 } }^{ 2 } }{ { { v }_{ 2 } }^{ 2 } } =\dfrac { 25 }{ 9 }$

$\dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { 5 }{ 3 }$

A man slides down a light rope whose breaking strength

$\eta$ times the weight

$(\eta < 1)$ . The maximum acceleration of the man so that the rope just breaks is:-

0%
$g(1 - \eta)$
0%
$g(1 +\eta)$
0%
$g\eta$
0%
$\dfrac{g}{\eta}$

Explanation

Breaking strength

$=$

$-$ maximum tension

Given that

$T=\eta \times$ weight of man

$=\eta \times mg$

From Newton's law of motion

$mg-T=ma$

$mg-\eta mg=ma$

$mg\left( 1-\eta \right) =ma$

$a=\left( 1-\eta \right) g$

Given that,

$\eta <1$ . So, acceleration is downwards because

$\left( 1-\eta \right) >0$

So,

$a=\left( 1-\eta \right) g$ option (A) correct.

Two persona A and B are sitting on weighing machines arranged in two vans parked at the equatorial region. They show same reading. If A moves due west and B moves due east with a speed of 200 kmph along the equator, then

0%
the weights of A and B will be same
0%
the weight of A slightly increases
0%
the weight of B slightly increases
0%
the weight of A slightly decreases

Two planet have the same average density but their radii are

${ R }_{ 1 }$ and

${ R }_{ 2 }$ . If acceleration due to gravity on these planets be

${ g }_{ 1 }$ and

${ g }_{ 2 }$ . respectively, then

0%
$\frac { { g }_{ 1 } }{ g_{ 2 } } =\frac { { R }_{ 1 } }{ { R }_{ 2 } }$
0%
$\frac { { g }_{ 1 } }{ g_{ 2 } } =\frac { { R }_{ 2 } }{ { R }_{ 1 } }$
0%
$\frac { { g }_{ 1 } }{ g_{ 2 } } =\frac { { { R }_{ 1 }^{ 2 } } }{ { R }_{ 2 }^{ 2 } }$
0%
$\frac { { g }_{ 1 } }{ g_{ 2 } } =\frac { { { R }_{ 1 }^{ 3 } } }{ { R }_{ 2 }^{ 3 } }$

Two identical star of mass

$M$ each orbit around their center of mass with orbital speed

$v$ due to interaction force between them. The escape velocity of an mass m from the center of mass of two stars system is

$(M>>m)$

0%
$4v$
0%
$v$
0%
$v/2$
0%
$\sqrt2$

The depth at which the effective value of acceleration due to gravity is one fourth that of its value on the earth's surface is

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

The distance between two planets of masses M and 4M is 'a' what is the gravitational potential at a point on a line joining them of which the gravitational intensity is zero?

0%
$-\frac{9GM}{r}$
0%
$-\frac{5GM}{a}$
0%
$-\frac{3GM}{a}$
0%
$-\frac{7GM}{a}$

Explanation

Let the point be the position when the gravitational field in zero

$\dfrac{Gm}{x^{2}}=\dfrac{G(4m)}{(r-x)^{2}} \Rightarrow \dfrac{1}{x} =\dfrac{2}{(r-x)}$

$r-2=2x \ x= r/3$

The point

$'P'$ is at a distance

$\dfrac{r}{3}$ from man

$m$ and

$\dfrac{2r}{3}$ from man

$4m$

$\therefore v=\dfrac{-Gm}{r/3}-\dfrac{G(4m)}{\dfrac{2r}{3}}=\dfrac{-9Gm}{r}$

$v=\dfrac{-9Gm}{r}$

$g$ is acceleration due to gravity on the surface of the earth

$,$ having radius

$R,$ the height at which the acceleration due to gravity reduces to

$g/2$ is

0%
$R/2$
0%
$\sqrt 2 R$
0%
$\frac{R}{{\sqrt 2 }}$
0%
$\left( {\sqrt 2 - 1} \right)R$

A satellite is moving on a circular orbit around a planet. A sharp impulse is given to the satellite such that escapes from the gravitational field of the planet. The impulse can be given in two different ways ways as shown. The ratio of the impulse given in the two different cases

$\dfrac{P_{1}}{P_{2}}$ is

0%
$1$
0%
$2$
0%
$\sqrt{2}-1$
0%
$\sqrt{2}+1$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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