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Physics 2 Marks Questions

Gravitation · MP Board (MPBSE) STD 11 Science (Madhya Pradesh - English Medium). 86 questions.

Questions · Page 2 of 2

2 Marks Questions

Question 512 Marks
An astronaut, by mistake, drops his food packet from an artificial satellite orbiting around the earth. Will it reach the surface of the earth? Why?
Answer
The food packet will not fall on the earth. As the satellite as well as astronaut were in a state of weightlessness, hence, the food packet, when dropped by mistake, will also start moving with the same velocity as that of satellite and will continue to move along with the satellite in the same orbit.
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Question 522 Marks
Earth's radius is about $6370\ km.$ A mass of $20\ kg$ is taken to a height of $160\ km$ above the earth's surface.
  1. What is the mass of the objects at that height?
  2. How much does the object weigh at this height?
Answer
  1. The mass of the object remain $20\ kg$ at that height.
  2. $\text{W}=\frac{\text{GMm}}{\text{r}^2}$
$\Rightarrow\frac{\text{W}_2}{\text{W}_1}=\frac{\text{r}^2_1}{\text{r}^2_2}$
since $G, M$ and $m$ are constant.
$\therefore\text{W}_2=9.8\times20\times\Big(\frac{6370}{6370+160}\Big)^2=186.5\text{N}$
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Question 532 Marks
A mass $M_1$ revolves around another mass $M_2$ in a path of radius r, what is the angular momentum associated with $M_1$?
Answer
Angular momentum L is the product of twice the mass and aerial velocity. $\therefore\text{L}=2\text{M}\frac{\text{dA}}{\text{dt}}=2\text{M}_1\times\frac{\pi\text{r}^2}{\text{T}}$ $=2\text{M}_1\frac{\pi\text{r}^2\text{v}}{2\pi\text{r}}$ $=\text{M}_1\text{r}\sqrt{\frac{\text{GM}_2}{\text{r}}}$ $=\text{M}_1\sqrt{\text{GM}_2\text{r}}$
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Question 542 Marks
An astronaut, by mistake, drops his food packet from an artificial satellite orbiting around the Earth. Will it reach the surface of Earth? Why?
Answer
The food packet will not fall on the Earth. As the satellite as well as astronaut were in a state of weightlessness, hence the food packet, when dropped by mistake, will also start moving with the same velocity as that of satellite and will continue to move along with the satellite in the same orbit.
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Question 552 Marks
On what factor does the escape speed from a surface depend?
Answer
Value of escape speed at the surface of a planet is given by the relation, $\text{v}_\text{es}=\sqrt{\frac{2\text{GM}}{\text{R}}}=\sqrt{2\text{gR}}$ Thus, the value of escape speed from the surface of a planet depends upon (i) value of acceleration due to gravity gat the surface and (ii) the size (i.e. radius) R of the planet only. It is independent of all other factors. e.g. The mass and size of the body to be projected, angle of projection, etc.
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Question 562 Marks
Give one example each of central force and non - central force.
Answer
Gravitational force, electrostatic force due to point mass and point charges are the examples of central force. Spin dependent nuclear force, magnetic force between two current carrying loops are the examples of non - central forces.
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Question 572 Marks
The value of acceleration due to gravity at the moon is $\frac{1}{6}\text{th}$ of the value of g at the surface of the earth, and the diameter of the moon is $\frac{1}{4}\text{th}$ of the diameter of the earth. Compare the ratio of the escape velocities.
Answer
$\text{v}_{\text{e}}=\sqrt{\frac{2\text{GM}}{\text{R}}}=\sqrt{2\text{gR}}$ $=\sqrt{\text{gD}}$ $\frac{(\text{v}_{\text{e}})_{\text{moon}}}{(\text{v}_{\text{e}})_{\text{earth}}}=\frac{\sqrt{(\text{gD})_{\text{moon}}}}{\sqrt{(\text{gD})_{\text{earth}}}}$ $\sqrt{\frac{1}{6}\times\frac{1}{4}}=\frac{1}{4.9}$
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Question 582 Marks
A planet reduces its radius by 1% with its mass remaining same. How acceleration due to gravity varies?
Answer
When mass is same, $\text{g}\propto\frac{1}{\text{R}^2}.$$\therefore\frac{\Delta\text{g}}{\text{g}}=2\frac{\Delta\text{R}}{\text{R}}$
% variation of g is 2%.
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Question 592 Marks
A spacecraft consumes more fuel in going from earth to moon than it does on the return trip. Comment on this.
Answer
In going from earth to moon, the spacecraft has to do more work against the greater gravitational attraction of the earth. For the return journey, moon's gravitational force is much less, hence less work is done and less fuel is consumed.
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Question 602 Marks
What is the gravitational force on a body inside a spherical shell? Why is it so?
Answer
Inside a spherical shell, gravitational force is zero. Since there is no mass inside, the gravitational field and thereby the force is zero.
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Question 612 Marks
An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer
Astronaut inside a small spaceship experience a very small negligible constant acceleration and hence astronaut feel weight less ness. If the space station has too much large mass and size then he can only experience acceleration due to gravity in the proximity of moon or on the moon.
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Question 622 Marks
Why a tennis ball bounces higher on hills than on plains?
Answer
As the acceleration due to gravity on hills is less than that on the surface of the earth (effect of height), therefore, a tennis ball bounces higher on hills than on plains.
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Question 632 Marks
What is the height at which the value of g is the same as at a depth of $\frac{\text{R}}{2}?$
Answer
At depth $\frac{\text{R}}{2},\text{g}'=\text{g}\Big(1-\frac{\text{R}}{2\text{R}}\Big)=\frac{\text{g}}{2}$At height $\text{x},\text{g}'=\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)$
$\therefore\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)=\frac{\text{g}}{2}$
$\frac{1}{2}=\frac{2\text{x}}{\text{R}}$
$\therefore\text{x}=\frac{\text{R}}{4}$
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Question 642 Marks
A body has a sense of weightlessness in a satellite revolving around the earth, why?
Answer
The astronauts and the satellite require the centripetal force to revolve around the earth. Their weight is used up in providing the necessary centripetal force. Hence an astronaut feels weightlessness in the space.
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Question 652 Marks
State Newton's law of gravitation in vector form.
Answer
The force of attraction between a pair of masses $m_1$ and $m_2$ separated by a length 'r' is given by. $\overrightarrow{\text{F}_{12}}=-\frac{\text{Gm}_1\text{m}_2}{\text{r}_2}\hat{\text{r}}_{21}$ $[\text{F}_{12}\rightarrow\text{force on }1\text{ due to }2].$ $\hat{\text{r}}\rightarrow\text{Points from }2\text{ to }1.$ -ve sign shows that the force is attractive.
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Question 662 Marks
The moon takes about 27 days to complete one orbit around the earth. The orbit is nearly a circle of radius $3.8 \times 10^8 \mathrm{~m}$. Calculate the mass of the earth from this data.
Answer
Let M and m be the mass of the earth and the moon respectively. The gravitational force of attraction provides the centripetal force. $\frac{\text{GMm}}{\text{r}^2}=\frac{\text{mv}^2}{\text{r}}$ $\Rightarrow\text{M}=\frac{\text{v}^2\text{r}}{\text{G}}=\frac{\omega^2\text{r}^3}{\text{G}}$ $=\Big(\frac{2\pi}{27\times24\times3600}\Big)^2\times\frac{(3.8\times10^8)^3}{6.67\times10^{-11}}$ $=5.968\times10^{24}\text{kg}$
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Question 672 Marks
A rocket is fired with a velocity 0.6 times the escape velocity on the surface of earth. How high will it go from the surface?
Answer
Velocity provided $= 0.6v_e$, $\text{v}_{\text{e}}=\sqrt{\frac{2\text{GM}}{\text{R}}}$ Applying conservation, $-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{m}(6\text{v}_{\text{e}})^2=-\frac{\text{GMm}}{\text{R}+\text{h}}$ Solve for h to get $\text{h}=\frac{72}{28}\text{R}.$
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Question 682 Marks
Answer the following: An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer
Yes, Astronaut can hope to detect gravity if the size of th e spaceship is extremely large, then the magnitude of the gravity will become appreciable and hence the gravitational effect of the spaceship may been me measurable.
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Question 692 Marks
If a planet existed whose mass and radius were both half of those of the earth, what would be the value of the acceleration due to gravity on its surface as compared to what it is on the earth's surface?
Answer
We know, $\text{g}=\frac{\text{GM}}{\text{R}^2}\dots(1)$ $\text{g}'=\frac{\text{GM}'}{(\text{R}')^2}\Big[\text{M}'=\frac{1}{2}\text{M,}\text{R}'=\frac{1}{2}\text{R}\Big]\dots(2)$ Form (1) and (2), we have $\frac{\text{g}'}{\text{g}}=\frac{1}{2}2^2=2$ $\therefore\text{g}'=2\text{g}$
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Question 702 Marks
Show graphically how g varies as you move from the centre of earth to great heights above the surface.
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Question 712 Marks
A mass M is broken into two parts, m and ( M - m ). How is m related to M so that the gravitational force between two parts is maximum?
Answer
Let $m_1 = m, m_2 = M - m \text{F}=\text{G}=\frac{\text{m}(\text{M}-\text{M})}{\text{r}^2}=\frac{\text{G}}{\text{r}}(\text{Mm}-\text{m}^2)$
Differentiating w.r.t. $\text{m},\frac{\text{dF}}{\text{dm}}=\frac{\text{G}}{\text{r}^2}(\text{M})-2\text{m})$
For F to be maximum, $\frac{\text{dF}}{\text{dm}}=0$
$\frac{\text{G}}{\text{r}^2}(\text{M}-2\text{m})=0$
$\text{M}=2\text{m},$ or $\text{m}=\frac{\text{M}}{2}$
$\therefore\text{m}_1=\text{m}_2=\frac{\text{M}}{2}$
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Question 722 Marks
Two particles of equal mass 'm' go round a circle of radius R, under the action of their mutual gravitational attraction. What is the speed of each particle?
Answer

The gravitational force of attraction provides the necessary centripetal force.
$\frac{\text{mv}^2}{\text{R}}=\frac{\text{G}(\text{m})(\text{m})}{(2\text{R})^2}$
$\Rightarrow\text{v}^2=\frac{\text{Gm}}{4\text{R}}$
$\Rightarrow\text{v}=\frac{1}{2}\sqrt{\frac{\text{Gm}}{\text{R}}}$
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Question 732 Marks
A black hole is a body from whose surface nothing may ever escape. What is the condition for a uniform spherical mass M to be a black hole? What should be the radius of such a black hole if its mass is the same as that of the Earth?
Answer
For a body to be a black hole, the escape velocity should be such that even light cannot escape. The limiting case for escape velocity is, $\sqrt{\frac{2\text{GM}}{\text{R}}}\leq\text{c}(\text{speed of light})$ For our earth, $M = M_e = 6 \times 10^{24}kg$, $\text{R}=\frac{2\text{GM}}{\text{c}^2}=9\times10^{-2}\text{m or }9\text{cm}$
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Question 742 Marks
Does the concentration of the earth's mass near its centre change the variation of g with height compared with a homogeneous sphere, how?
Answer
Any change in the distribution of the earth's mass will not affect the variation of acceleration due to gravity with height. This is because for a point outside the earth, the whole mass of the earth is effective and the earth behaves as a homogeneous sphere.
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Question 752 Marks
If earth be at half of its present distance from the sun, how many days will there be in a year?
Answer
$\text{T}^2\propto\text{r}^3,\text{Since r}\rightarrow\frac{\text{r}}{2},\frac{\text{T}^2_1}{\text{T}^2}=\Big(\frac{1}{2}\Big)^3$ $\therefore\text{T}_1=\Big(\frac{1}{8}\Big)^{\frac{3}{2}}\text{T};\text{T}_1=\Big(\frac{1}{2\sqrt{2}}\Big)^3\text{T}$ $\text{i.e.}\text{T}_1=\frac{\text{T}}{16\sqrt{2}}$
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Question 762 Marks
Why are space crafts usually launched from west to east? Why is it more advantageous to launch rockets in the equatorial plane?
Answer
Earth rotates on its axis from west to east. A satellite launched from west to east will have the advantage of the additional velocity of the earth's rotation. The effect is maximum at the equator, hence it is most advantageous to launch the satellite from west to east on the equatorial plane.
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Question 772 Marks
Answer the following: You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Answer
No. Electrical forces depend upon the nature of the intervening medium while the gravitational forces don’t depend upon the nature of the intervening medium. So, such shielding acts are not possible in case of gravitation i.e., gravity screens are not possible.
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Question 782 Marks
Answer the following: If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?
Answer
Earth moon distance is very small as compared to earth-sun distance. Tidal effect is inversely proportional to the cube of the distance it means it is not governed by inverse square law like the gravitational force (which obeys inverse square law). Hence, tidal effect of moon is larger than that due to the sun.
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Question 792 Marks
Define gravitational potential at a point.
Answer
The work done in carrying unit mass from infinity to a point in gravitational field is gravitational potential. $\text{V}_{\text{g}}=-\frac{\text{GM}}{\text{r}}$ where r is the distance of the point from M.
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Question 802 Marks
Two satellites A and B go around a planet Pin circular orbits having radius 4R and R respectively. If the speed of the satellite A is 3V, find the speed of the satellite B.
Answer
$\text{As},\text{v}_{\text{o}}=\sqrt{\frac{\text{GM}}{\text{R}}}$ $\text{So},3\text{v}=\sqrt{\frac{\text{GM}}{4\text{R}}}\text{ and }\text{v}'=\sqrt{\frac{\text{GM}}{\text{R}}}$ $\therefore\frac{\text{v}'}{3\text{v}}=2\text{ or }\text{v}'=6\text{v}$
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Question 812 Marks
If you compare the gravitational force on the earth due to the sun to that of due to the moon, you would find that the sun's pull is greater than the moon's pull. However, the tidal effect of the moon's pull is greater than the tidal effect of the sun. Why?
Answer
Whereas the gravitational force depends inversely on the square of the distance, tidal effect depends inversely on the cube of the distance.
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Question 822 Marks
How is the gravitational force between two point masses affected when they are dipped in water keeping the separation between them the same?
Answer
The Newton’s Universal law of gravitational force of attraction (F) between two bodies of masses $\mathrm{m}_1, \mathrm{~m}_2$ separated by distance r is $\text{F}=\frac{\text{Gm}_1\text{m}_2}{\text{r}^2}$ G does not depend upon the medium. So force of attraction does not change if the masses are kept in water or any other medium.
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Question 832 Marks
Objects at rest on the earth's surface move in circular paths with a period of 24 hours. Are they in orbit in the same sense that an earth's satellite is in orbit? Why not? What would the length of the day have to be to put such objects in a true orbit?
Answer
The objects on the earth's surface are not in orbital motion w.r.t. the earth. In order that an object has an orbital motion close to the earth's surface, its orbital velocity, and period of motion must be, $\text{v}_{\text{o}}=\sqrt{\text{gR}};$ $\text{T}=\frac{2\pi\text{R}}{\text{v}_{\text{o}}}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}=1.4\text{hrs.}$ Therefore, the length of the day has to be 1.4hrs in case the objects on the earth's surface are in the true orbital motion like that of the earth satellite.
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Question 842 Marks
An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer
In case the size of the space-station becomes large, the forces of gravity will become appreciable, and the astronaut can hope to detect it.
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Question 852 Marks
Shown are several curves. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).
Answer
The trajectory of a projectile under gravitational force of earth is a conic section or parabolic or elliptical or its part whose focus must be the centre of the earth. Only (c) option amongst the given curves show the centre of the earth as the focus of trajectory.
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Question 862 Marks
Why do different planets have different escape velocities?
Answer
Escape velocity, $\text{v}=\sqrt{2\text{gR}}=\sqrt{\frac{2\text{GM}}{\text{R}}}.$ Thus escape velocity of a planet depends upon (i) its mass (M) and (ii) its size (R). As different planets have different masses and sizes, so they have different escape velocities.
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2 Marks Questions - Page 2 - Physics STD 11 Science Questions - Vidyadip