MCQ 511 Mark
In our solar system, the interplanetary region has chunks of matter $($much smaller in size compared to planets$)$ called asteroids. They
- A
Will not move around the sun since they have very small masses compared to sun.
- ✓
Will move in an irregular way because of their small masses and will drift away into outer space.
- C
Will move around the sun in closed orbits but not obey Kepler’s laws.
- D
Will move in orbits like planets and obey Kepler’s laws.
AnswerCorrect option: B. Will move in an irregular way because of their small masses and will drift away into outer space.
Asteroids will move in orbits like planets and obey Kepler’s law because they are also being acted upon by central gravitational forces.
View full question & answer→MCQ 521 Mark
Average density of the earth:
- ✓
Is directly proportional to $g.$
- B
Is inversely proportional to $g.$
- C
Is a complex function of $g.$
- D
Does not depend on $g.$
AnswerCorrect option: A. Is directly proportional to $g.$
$\text{g}=\frac{\text{GM}}{\text{R}^2}\times\frac{4}{3}\pi\text{R}^3$
$\rho=\frac{4}{3}\pi\text{GR}\rho$
or $g\propto\rho$
View full question & answer→MCQ 531 Mark
If the mass of sun were ten times smaller and gravitational constant $G$ were ten times larger in magnitudes.
- Walking on ground would became more difficult.
- The acceleration due to gravity on earth will not change.
- Raindrops will fall much faster.
- Airplanes will have to travel much faster.
- A
$A$ and $B$
- B
$A , B$ and $D$
- C
$B$ and $A$
- ✓
$A , C$ and $D$
AnswerCorrect option: D. $A , C$ and $D$
If the gravitational constant $G$ becomes $10$ times larger in magnitude.
$\text{G}'=10\text{G}$
Gravitational field due to the earth
$\text{g}'=\frac{\text{G}'\text{M}_\text{e}}{\text{r}^2}$
$=\frac{10\text{GM}_\text{e}}{\text{r}^2}=10\text{g}$
Weight of a person $=\text{mg}'=\text{m}\times10\text{g}=10\text{mg}$
Force on the man due to sun, $\text{F}=\frac{\text{GM}'_\text{s}\text{m}}{\text{r}^2}$
Mass of the sun $\text{M}'_\text{s}=\frac{1}{10}\text{M}_\text{s}\Rightarrow10\text{M}'_\text{s}=\text{M}_\text{s}$
$\text{F}=\frac{\text{GM}_\text{s}\text{m}}{\text{10r}^2}$
Weight of person becomes $10$ times larger so it will be more difficult to walk. Option $(a)$ is correct.
As $g’ = 10g,$ the acceleration due to gravity changes. Option $(b)$ is incorrect.
The terminal velocity $\text{v}_\text{T}\propto\text{g}$ and $g', g' = 10g,$ the terminal velocity increases $10$ times.
Hence the rain drops falls $10$ times faster. Option $(c)$ is correct.
As the $g’ = 10g,$ to overcome the increase gravitational force and in order to maintain the speed the aeroplane will have to travel much faster. Option $(d)$ is correct.
View full question & answer→MCQ 541 Mark
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 earth to a height equal to the radius $R$ of the earth is:
AnswerCorrect option: A. $\frac{\text{I}}{2}\text{mgR}$
Work done in raising the body
$=\int^\limits{2\text{R}}_\limits{\text{R}}\frac{\text{GMm}}{\text{x}^2}\text{dx}$
$=\int^\limits{2\text{R}}_\limits{\text{R}}\frac{\text{gR}^2}{\text{x}^2}\text{mdx}$
$=\frac{1}{2}\text{mgR}$
View full question & answer→MCQ 551 Mark
A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a,$ then the time period is given by $\text{t}=2\pi\sqrt{\frac{1}{\text{g}}}$ where $g'$ is equal to:
AnswerCorrect option: D. $\sqrt{\text{g}^2+\text{a}^2}$
Bob of the pendulum is under the effect of two forces.
Weight $= mg$ acting ertically downwards.
Horizontal force $= ma.$
Therefore, resultant force $=\sqrt{(\text{gm})^2+(\text{mg})^2}$
Hence, acceleration of the bob
$=\sqrt{\text{g}^2+\text{a}^2}$
View full question & answer→MCQ 561 Mark
The orbital speed of an artificial satellite in a circular orbit just above the earth's surface is $v$. For a satellite orbiting at an altitude of half the earth's radius the orbital speed is:
- A
$\Big(\frac{3}{2}\Big)\text{v}$
- B
$\Big(\sqrt{\frac{3}{2}}\Big)\text{v}$
- ✓
$\Big(\sqrt{\frac{2}{3}}\Big)\text{v}$
- D
$\Big(\frac{2}{3}\Big)\text{v}$
AnswerCorrect option: C. $\Big(\sqrt{\frac{2}{3}}\Big)\text{v}$
Orbital velocity close to earth, $\text{v}=\sqrt{\text{gR}}$ Orbital velocity at height $\frac{\text{R}}{2}$ will be
$\text{v}=\sqrt{\frac{\text{gR}^2}{\text{R}+\frac{\text{R}}{2}}}$
$=\sqrt{\frac{2}{3}\text{gR}}$
$=\Big(\sqrt{\frac{2}{3}}\Big)\text{v}$
View full question & answer→MCQ 571 Mark
Consider a planet moving in an elliptical planet orbit round the sun. The work done on the planet by the gravitational force of the sun:
- A
Is zero in some part of the orbit.
- B
Is zero in no part of the motion.
- C
Is zero in any small part of the orbit.
- ✓
Is zero in one complete revolution.
AnswerCorrect option: D. Is zero in one complete revolution.
View full question & answer→MCQ 581 Mark
A satellite is orbiting the earth. If its distance from the earth is increased, its:
- A
Angular velocity would increase.
- B
Linear velocity would increase.
- C
Angular velocity would decrease.
- ✓
Time period would increase.
AnswerCorrect option: D. Time period would increase.
View full question & answer→MCQ 591 Mark
The gravitational potential at a distance $r$ from the centre of the earth $(r > R)$ is given by $($consider, mass of the earth $= M,$ radius of the earth $= R):$
- A
$\frac{-\text{GM}_\text{e}}{\text{R}}$
- B
$\frac{\text{GM}_\text{e}}{\text{R}}$
- ✓
$\frac{-\text{GM}_\text{e}}{\text{r}}$
- D
$\frac{+\text{GM}_\text{e}}{\text{r}}$
AnswerCorrect option: C. $\frac{-\text{GM}_\text{e}}{\text{r}}$
The gravitational potential at a distance from the centre of the earth.
$\text{v(r)}=-\frac{\text{GM}_\text{e}}{\text{r}}$
View full question & answer→MCQ 601 Mark
Two point masses $m_1$ and $m_2$ are separated by a distance $r.$ The gravitational potential energy of the system is $G_1$. When the separation between the particles is doubled, the gravitational potential energy is $G_2$. Then, the ratio of $\frac{\text{G}_1}{\text{G}_2}$ is:
View full question & answer→MCQ 611 Mark
For a satellite in elliptical orbit which of the following quantities does not remain constant?
AnswerIn elliptical orbit, velocity keeps on changing both in magnitude and direction.
Therefore, momentum does not remain constant $(P = mv).$
View full question & answer→MCQ 621 Mark
The period of revolution of a certain planet in an orbit of radius $R$ is $T.$ Its period of revolution in an orbit of radius $4R$ will be:
- A
$2\text{T}$
- B
$\frac{2}{2\text{T}}$
- C
$4\text{T}$
- ✓
$8\text{T}$
AnswerCorrect option: D. $8\text{T}$
$\frac{\text{T}^2_2}{\text{T}^2_1}=\frac{\text{r}^3_2}{\text{r}^2_1}$
or $\text{T}_2=\text{T}_1\Big(\frac{\text{r}_2}{\text{r}_1}\Big)^{\frac{3}{2}}$
$=\text{T}\Big(\frac{4\text{R}}{\text{R}}\Big)^{\frac{3}{2}}$
$=8\text{T}$
View full question & answer→MCQ 631 Mark
Which of the following statements is correct about satellites?
- A
A satellite cannot move in a stable orbit in a plane passing through the earth's centre.
- ✓
Geostationary satellites are launched in the equatorial plane.
- C
We can use just one geostationary satellite for global communication around the globe.
- D
The speed of satellite increases with an increase in the radius of its orbit.
AnswerCorrect option: B. Geostationary satellites are launched in the equatorial plane.
View full question & answer→MCQ 641 Mark
Earth is flattened at the poles and bulges at the equator. This is due to the fact that:
- A
The earth revolves around the sun in an elliptical orbit.
- B
The angular velocity of spinning about its axis is more at the equator.
- ✓
The centrifugal force is more at the equator than at poles.
- D
AnswerCorrect option: C. The centrifugal force is more at the equator than at poles.
Higher centrifugal force causes bulging of earth at equator.
View full question & answer→MCQ 651 Mark
If $g$ denotes the value of acceleration due to gravity at a point distant $r$ from the centre of earth of radius $R.$ If $r < R,$ then:
- A
$\text{g}\propto\text{r}^2$
- ✓
$\text{g}\propto\text{r}$
- C
$\text{g}\propto\frac{1}{\text{r}^2}$
- D
$\text{g}\propto\frac{1}{\text{r}}$
AnswerCorrect option: B. $\text{g}\propto\text{r}$
$\text{g}'=\text{g}\Big(1-\frac{\text{d}}{\text{g}}\Big)=\text{g}\Big(\frac{\text{R}-\text{d}}{\text{R}}\Big)$
$=\text{g}\frac{\text{r}}{\text{g}},$
i.e.$\text{ g}'\propto\text{r}.$
View full question & answer→MCQ 661 Mark
The mean radius of the earth is $R,$ its angular speed about its own axis is $o$ and the acceleration due to gravity at earth surface is $g.$ The cube of radius of orbit of 'geostationary satellite' will be:
- A
$\Big(\frac{\text{R}^2\text{g}}{\omega}\Big)$
- B
$\Big(\frac{\text{R}^2\omega}{\text{g}}\Big)$
- C
$\Big(\frac{\text{R}\text{g}}{\omega^2}\Big)$
- ✓
$\Big(\frac{\text{R}^2\text{g}}{\omega^2}\Big)$
AnswerCorrect option: D. $\Big(\frac{\text{R}^2\text{g}}{\omega^2}\Big)$
$\frac{\text{GMm}}{\text{r}^2}=\text{mr}\omega^2$
or $\text{r}^3=\frac{\text{GM}}{\omega^2}=\frac{\text{gR}^2}{\omega}^2$
$\Big(\because\text{g}=\frac{\text{GM}}{\text{R}^2}\Big)$
View full question & answer→MCQ 671 Mark
Suppose universal gravitational constant starts to decrease, then:
- A
Length of the day, on earth, will decrease.
- B
Kinetic energy of earth will decrease.
- C
Earth will follow a spiral path of increasing radius.
- ✓
View full question & answer→MCQ 681 Mark
The time period of an earth satellite in circular orbit is independent of:
- ✓
The mass of the satellite.
- B
- C
Both the mass of satellite and radius of the orbit.
- D
Neither the mass of satellite nor the radius of the orbit.
AnswerCorrect option: A. The mass of the satellite.
Time period of satellite at height $h$ ground is
$\text{T}=\frac{2\pi}{\text{R}}\sqrt{\frac{(\text{R+h}^3)}{\text{g}}}$
$=\frac{2\pi\big(\text{R+h}\big)^{\frac{3}{2}}}{(\text{GM})^{\frac{1}{2}}}$
It is independent of mass $m$ of the satellite.
View full question & answer→MCQ 691 Mark
If the sun and the planets carried huge amounts of opposite charges,
- All three of Kepler’s laws would still be valid.
- Only the third law will be valid.
- The second law will not change.
- The first law will still be valid.
- A
$A$ and $B$
- ✓
$A , C$ and $D$
- C
$B$ and $D$
- D
AnswerCorrect option: B. $A , C$ and $D$
Due to opposite charges the attractive electrostatic forces of attraction produced will be large, along with the gravitational forces. Both the forces will be added and would be radial in nature. Since both the forces are central forces and obey the inverse square law, all the three kepler's laws will be valid.
View full question & answer→MCQ 701 Mark
The centre of mass of an extended body on the surface of the earth and its centre of gravity.
AnswerCorrect option: D. Is close to each other for objects, say of sizes less than $100m.$
The centre of gravity $\text{(COG)}$ is based on weight of a body. The centre of mass $\text{(COM)}$ is based on mass of a body. The $\text{COG}$ is a point in a body over which the body can be perfectly balanced. The net torque due to gravity at $\text{COG}$ is zero. Whereas the $\text{COM}$ is the average location of the mass distribution of a body or it is a point where the whole mass of the body is supposed to be concentrated. If given some angular momentum the body would spin about the $\text{COM.}$ For small body, say of size less than $100m, << R_e$ when placed in uniform gravitational field, the $\text{COM}$ is very close to $\text{COG.}$ If the size of the body increases i.e., very larger like lake, or mountain, its weight changes, and its $\text{COM}$ and $\text{COG}$ become far from each other. As in case of a spherical ball where the $\text{COM}$ and $\text{COG}$ are same, but in case of a mountain the $\text{COM}$ and the $\text{COG}$ are not aligned, where $\text{COM}$ lies a bit above its $\text{COG}$. Option $(d)$ is justified.
View full question & answer→MCQ 711 Mark
Choose the correct statement/ (s):
- A
Weight of a body is greater on planes and less on hill tops.
- B
Weight of a body is greater at the poles and less at the equator.
- C
Weight of a body on the moon is less than that on the earth.
- ✓
AnswerExplanation:
Weight = Mass × Acceleration due to gravity. The acceleration due to gravity is greater on planes than on hill top. g is greater at poles than at the equator. g is less on moon than on the earth.
View full question & answer→MCQ 721 Mark
Three uniform spheres of mass $M$ and radius $R$ each are kept in such a way that each touches the other two. The magnitude of the gravitational force on any of the spheres due to the other two is:
- ✓
$\frac{\sqrt{3}}{4}\frac{\text{GM}^2}{\text{R}^2}$
- B
$\frac{3}{2}\frac{\text{GM}^2}{\text{R}^2}$
- C
$\frac{\sqrt{3}\text{GM}^2}{\text{R}^2}$
- D
$\frac{\sqrt{3}}{2}\frac{\text{GM}^2}{\text{R}^2}$
AnswerCorrect option: A. $\frac{\sqrt{3}}{4}\frac{\text{GM}^2}{\text{R}^2}$
View full question & answer→MCQ 731 Mark
The gravitational potential energy of a system consisting of two particles separated by a distance $r$ is:
- A
Directly proportional to product of the masses of particles.
- B
Inversely proportional to the separation betweenthem.
- C
- ✓
Both $(a)$ and $(b).$
AnswerCorrect option: D. Both $(a)$ and $(b).$
View full question & answer→MCQ 741 Mark
A body of mass $m$ is placed on earth surface which is taken from earth surface to a height of $h = 3R,$ then change in gravitational potential energy is:
- A
$\frac{1}{4}\text{mgR}$
- B
$\frac{2}{3}\text{mgR}$
- ✓
$\frac{3}{4}\text{mgR}$
- D
$\frac{1}{3}\text{mgR}$
AnswerCorrect option: C. $\frac{3}{4}\text{mgR}$
Change in gravitational potential energy
$=\frac{-\text{GMm}}{(3\text{R}+\text{R})}-\Big(\frac{-\text{GMm}}{\text{R}}\Big)$
$=\frac{3}{4}\frac{\text{GMm}}{\text{R}}$
$=\frac{3}{4}\text{mgR}.$
View full question & answer→MCQ 751 Mark
What is the value of acceleration caused by force of gravity on a stone placed on ground?
MCQ 761 Mark
The radius of the orbit of a satellite is $r$ and its kinetic energy is $K.$ If the radius of the orbit is doubled, then the new kinetic energy $K'$ is:
- A
$2$
- ✓
$\frac{\text{K}}{2}$
- C
$4K$
- D
AnswerCorrect option: B. $\frac{\text{K}}{2}$
View full question & answer→MCQ 771 Mark
Two spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F.$ The space around the masses is now filled with liquid of specific gravity $3$. The gravitational force will now be:
- A
$3\text{F}$
- ✓
$\text{F}$
- C
$\frac{\text{F}}{3}$
- D
$\frac{\text{F}}{3}$
AnswerCorrect option: B. $\text{F}$
The gravitational force between two bodies is independent of the presence of other bodies.
View full question & answer→MCQ 781 Mark
The law of areas can be interpreted as:
- ✓
$\frac{\Delta\text{A}}{\Delta\text{t}}=\text{constant}$
- B
$\frac{\Delta\text{A}}{\Delta\text{t}}=\frac{\text{L}}{\text{m}}$
- C
$\frac{\Delta\text{A}}{\Delta\text{t}}=\frac{1}{2}(\text{r}\times\text{P})$
- D
$\frac{\Delta\text{A}}{\Delta\text{t}}=\frac{2\text{L}}{\text{m}}$
AnswerCorrect option: A. $\frac{\Delta\text{A}}{\Delta\text{t}}=\text{constant}$
View full question & answer→MCQ 791 Mark
If three uniform spheres, each having mass $M$ and radius $r,$ are kept in such a way that each touches the other two, the magnitude of the gravitational force on any sphere due to the other two is:
- A
$\frac{\text{GM}^2}{4\text{r}^2}$
- B
$\frac{2\text{GM}^2}{\text{r}^2}$
- C
$\frac{2\text{GM}^2}{4\text{r}^2}$
- ✓
$\frac{\sqrt{3}\text{GM}^2}{4\text{r}^2}$
AnswerCorrect option: D. $\frac{\sqrt{3}\text{GM}^2}{4\text{r}^2}$
View full question & answer→MCQ 801 Mark
A satellite is launched into a circular orbit of radius $R$ around the earth. A second satellite launched into an orbit of radius $1.01R.$ The time period of the second satellite is larger than that of the first one by approximately:
- A
$0.5\%$
- ✓
$1.5\%$
- C
$1\%$
- D
$3.0\%$
AnswerCorrect option: B. $1.5\%$
View full question & answer→MCQ 811 Mark
The radii of two planets are respectively $R_1$ and $R_2$ and their densities are respectively $P_1$ and $P_2$. The ratio of the accelerations due to gravity at their surfaces:
- A
$\text{g}_1:\text{g}_2=\frac{\rho_1}{\text{R}^2_1}:\frac{\rho_2}{\text{R}^2_2}$
- B
$\text{g}_1:\text{g}_2=\text{R}_1\text{R}_2:\rho_1\rho_2$
- C
$\text{g}_1:\text{g}_2=\text{R}_1\rho_2:\text{R}_2\rho_1$
- ✓
$\text{g}_1:\text{g}_2=\text{R}_1\rho_1:\text{R}_2\rho_2$
AnswerCorrect option: D. $\text{g}_1:\text{g}_2=\text{R}_1\rho_1:\text{R}_2\rho_2$
View full question & answer→MCQ 821 Mark
The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity.
- A
Will be directed towards the centre but not the same everywhere.
- B
Will have the same value everywhere but not directed towards the centre.
- C
Will be same everywhere in magnitude directed towards the centre.
- ✓
Cannot be zero at any point.
AnswerCorrect option: D. Cannot be zero at any point.
Acceleration due to gravity $g = 0,$ at the centre if we assume the earth as a sphere of uniform density, then it can be treated as point mass placed at its centre.
But on surface of the earth the acceleration due to gravity cannot be zero at any point.
View full question & answer→MCQ 831 Mark
If a particle is fired vertically upwards from the surface of earth and reaches a height of $6400\ km,$ the initial velocity of the particle is $($assume $R = 6400\ km$ and $g = 10\ ms^{-2})$
- A
$4\ km/ \sec$
- B
$2\ km/ \sec$
- ✓
$8\ km/ \sec$
- D
$16\ km/ \sec$
AnswerCorrect option: C. $8\ km/ \sec$
View full question & answer→MCQ 841 Mark
A point mass $m$ is placed at the centre of the square $\text{ABCD}$ of side a units as shown below.
The resultant gravitational force on mass $m$ due to masses $m_1$ and $m_2$ plant on the vertices of square is: - A
$\frac{\text{Gm}_1\text{m}_2}{(\text{a}\sqrt{2})^2}$
- B
$\frac{2\text{Gm(m}_1+\text{m}_2)}{\text{a}^2}$
- ✓
$\text{zero}$
- D
$\frac{\text{Gm(m}_1+\text{m}_2)}{(\text{a}\sqrt{2})^2}$
AnswerCorrect option: C. $\text{zero}$
View full question & answer→MCQ 851 Mark
The radii of circular orbits of two satellites around the earth are in the ratio $1 : 4,$ then ratio of their respective periods of revolution is:
- A
$1 : 4$
- B
$4 : 1$
- ✓
$1 : 8$
- D
$8 : 1$
AnswerCorrect option: C. $1 : 8$
$\text{T}^2\propto\text{R}^3$
View full question & answer→MCQ 861 Mark
Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,
- A
The solar cells and batteries in satellites run out.
- B
The laws of gravitation predict a trajectory spiralling inwards.
- ✓
Of viscous forces causing the speed of satellite and hence height to gradually decrease.
- D
Of collisions with other satellites.
AnswerCorrect option: C. Of viscous forces causing the speed of satellite and hence height to gradually decrease.
The $P.E.$ of satellite orbiting in orbit of radius $r$ due to earth of mass $M$ is $\Big(\frac{-\text{GM}}{2\text{r}}\Big)$ negative sign shows the force of attraction between satellite and earth. Energy $(P.E.)$ is continuously reduced due to atmospheric friction, the radius of the orbit or height decreases gradually, and ultimately it comes back to earth with increasing speed and burns in the atmosphere.
View full question & answer→MCQ 871 Mark
The velocity of the planet when it is closest to sun is:
AnswerFrom conservation of angular momentum,
Velocity of planet $(v) \propto\frac{1}{\text{Distance of the planet from sun (r)}}$
So, $r_p$ is minimum for perihelion $(P).$
$= V_p$ is maximum.
View full question & answer→MCQ 881 Mark
Supposing Newton’s law of gravitation for gravitation forces $F_1$ and $F_2$ between two masses $m_1$ and $m_2$ at positions $r_1$ and $r_2$ read $\text{F}_1=-\text{F}_2=-\frac{\text{r}^{12}}{\text{r}^3_{12}}\text{GM}_0^2\Big(\frac{\text{m}_1\text{m}_2}{\text{M}^2_0}\Big)^\text{n}$ where $M_0$ is a constant of dimension of mass, $r_{12} = r_1 – r_2$ and $n$ is a number. In such a case,
AnswerAccording to the problem,
$\text{F}_1=-\text{F}_2=-\frac{\text{r}^{12}}{\text{r}^3_{12}}\text{GM}_0^2\Big(\frac{\text{m}_1\text{m}_2}{\text{M}^2_0}\Big)^\text{n}$
$\Rightarrow\ \vec{\text{r}}_{12}=\text{r}_1-\text{r}_2$
$(\text{a})\text{F}=\frac{\text{GM}^{2(1-\text{n})}_0(\text{M})^\text{n}}{\text{r}^2_{12}}(\text{m}_1\text{m}_2)^\text{n}$
Take, $m_1 = M ($mass of earth$), m_2 = m ($mas of the object), $r_{12} R ($radius of earth$)$
Therefore, $\text{F}=\Big(\frac{\text{GM}_0^{(2-2\text{n})}(\text{M})^\text{n}}{\text{R}^2}\Big)\text{m}^\text{n}=\text{Km}^\text{n}$
Where $K$ is the constant or the term in the bracket is regarded as constant.
As $\text{F}=\text{mg},$
so $g=\text{K m}^\text{n-1}$,
hence $g$ depends upon the mass of object.
Since, $g$ depends upon position vector and mass of object,
hence it will be different for different objects. As $g$ is not constant,
hence constant of proportionality will not be constant in Kepler’s third law.
Hence, Kepler’s third law will not be valid.
As the force is of central nature,
$\Big[\because\text{Force}\propto\frac{1}{\text{r}^2}\Big]$
Hence, first two Kepler’s laws will be valid.
Hence option $(b)$ in incorrect and $(c)$ is correct.
$(d)$ When $n$ is negative, $F = K / M^n$ Or we can say that $F$ is inversely proportional to mass. This implies that lighter bodies will experience a greater force than the heavier bodies and vice versa.
Hence, object lighter than water will sink in water.
View full question & answer→MCQ 891 Mark
If the radius of earth were to increase by $1\%$, its mass remaining the same, the acceleration due to gravity on the surface of earth will:
- A
Increase by $1\%$
- ✓
Decrease by $2\%$
- C
Decrease by $1\%$
- D
Increase by $2\%$
AnswerCorrect option: B. Decrease by $2\%$
$\text{g}=\frac{\text{GM}}{\text{R}^2},\frac{\Delta\text{g}}{\text{g}}$
$=-2\frac{\Delta\text{R}}{\text{R}}$
$\therefore$ Percentage change in the value of $g$
$=\frac{\Delta\text{g}}{\text{g}}\times100$
$=-2\frac{\Delta\text{R}}{\text{R}}\times100$
$=-2(1\%)$
$=-2\%$
View full question & answer→MCQ 901 Mark
Which one of the following statements is correct?
- A
The energy required to rocket an orbiting satellite out of earth's gravitational influence is more than the energy required to project a stationary object at the same height $($as the satellite$)$ out of earth's influence.
- B
If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of potential energy.
- C
The first artificial satellite sputnik I was launched in the year $2001.$
- ✓
The time period of rotation of the $\text{SYNCOMS.} ($Synchronous communications satellite$)$ is $24$ hours.
AnswerCorrect option: D. The time period of rotation of the $\text{SYNCOMS.} ($Synchronous communications satellite$)$ is $24$ hours.
View full question & answer→MCQ 911 Mark
A satellite is orbiting just above the surface of a planet of average density $\rho$ with period $T.$ If $G$ is the universal gravitational constant, the quantity $\text{T}^2\rho$ is equal to:
AnswerCorrect option: C. $4\frac{\pi}{\text{G}}$
$\text{v}=\sqrt{\text{gR}}$ and $T=\frac{2\pi\text{R}}{\text{v}}=\frac{2\pi\text{R}}{\sqrt{\text{gR}}}$
or $\text{T}=\frac{2\pi\text{R}}{\sqrt{\frac{\text{GM}}{\text{R}}}}=2\pi$
$\sqrt{\frac{\text{R}^3}{\text{G}\frac{4}{3}\pi\text{R}^3\rho}}$
$\text{T}^2\rho=4\pi^2\times\frac{3}{4\text{G}\pi}=\frac{3\pi}{\text{G}}$
View full question & answer→MCQ 921 Mark
There have been suggestions that the value of the gravitational constant $G$ becomes smaller when considered over very large time period $($in billions of years$)$ in the future. If that happens, for our earth,
- A
- B
After sufficiently long time we will leave the solar system.
- C
We will be going around but not strictly in closed orbits.
- ✓
Both $B$ and $C$
AnswerCorrect option: D. Both $B$ and $C$
We know that gravitational force exists between the earth and the sun.
$\text{F}_\text{G}=\frac{\text{G}(\text{M}_\text{S}\times\text{m}_\text{e})}{\text{r}^2}$ Where $M_S$ is mass of the sun and $m_e$ is mass of the earth.
This provides the necessary centripetal force for the circular orbit of the earth around the sun. As $G$ decreases with time, the gravitational force $F_G$ will become weaker with time. As $F_G$ is changing with time due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker.
Hence, after long time the earth will leave the solar system.
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Which of the following are true?
- A
A polar satellite goes around the earth’s pole in northsouth direction.
- B
A geostationary satellite goes around the earth in eastwest direction.
- C
A geostationary satellite goes around the earth in westeast direction.
- ✓
Both $A$ and $C$
AnswerCorrect option: D. Both $A$ and $C$
The satellite which appears stationary relative to earth is called the geostationary satellite. It revolves around the earth in the west$-$east direction with the same angular velocity as done by the earth about its own axis in the west$-$east direction. A polar satellite revolves around the earth's pole in north$-$south direction. It is independent of earth's rotation. Option $(a), (c)$ are correct.
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An object is thrown from the surface of the moon. The escape speed for the object is:
- ✓
$\sqrt{2\text{g}'\text{ R}_\text{m}},$ where $g' =$ acceleration due to gravity on the moon and $R_m =$ redius of the moon.
- B
$\sqrt{2\text{g}'\text{ R}_\text{e}},$ where $g' =$ acceleration due to gravity on the moon and $R_e =$ radius of the earth.
- C
$\sqrt{2\text{g}\text{ R}_\text{m}},$ where $g =$ acceleration due to gravity on the earth and $R_m =$ raduis of the moon.
- D
AnswerCorrect option: A. $\sqrt{2\text{g}'\text{ R}_\text{m}},$ where $g' =$ acceleration due to gravity on the moon and $R_m =$ redius of the moon.
Escape speed from the moon $=\sqrt{2\text{g}'\text{ R}_\text{m}}$ where,
$g' =$ acceleration due to gravity on the surface of moon.
$R_m =$ radius of the moon.
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The escape speed from the surface of earth is $v_e$. The escape speed from the surface of a planet whose mass and radius are $3$ times those of the earth will be:
- ✓
$v_e$
- B
$3v_e$
- C
$9v_e$
- D
$27v_e$
Answer$\text{v}_{\text{e}}=\sqrt{\text{vgR}}$
$=\sqrt{\frac{2\text{GM}_{\text{p}}}{\text{R}_{\text{p}}}}$
$=\sqrt{\frac{2\text{G}\times3\text{M}}{2\text{R}}}$
$=\sqrt{\frac{2\text{GM}}{\text{R}}}$
$=\text{v}_{\text{e}}.$
View full question & answer→MCQ 961 Mark
Different points in earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the $c.m. ($centre of mass$)$ causing translation and a net torque at the $c.m.$ causing rotation around an axis through the $c.m.$ For the earthsun system $($approximating the earth as a uniform density sphere$).$
- ✓
- B
The torque causes the earth to spin.
- C
The rigid body result is not applicable since the earth is not even approximately a rigid body.
- D
The torque causes the earth to move around the sun.
AnswerAs the earth is revolving around the sun in a circular motion $($approximately in actual the path of earth around the sun is elliptical$)$ due to gravitational attraction.
When we consider the earth$-$sun as a single system and we are taking earth as a sphere of uniform density.
Then the gravitational force $(F)$ will be of radial nature,
i.e. angle between position vector $r$ and force $F$ is zero.
So, torque
$|\vec{\tau}|=|\vec{\text{r}}\times\vec{\text{F}}|=\text{r F}\sin0^0=0.$
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