If F<\/em> is the gravitational force acting on a body of mass m<\/em>, then a<\/em> is the acceleration of a free falling body.<\/p>\n\n\n\n
This force is given as<\/p>\n\n\n\n
F=GMmR2.
Here, M is the mass of the Earth; G is the universal gravitational constant and R is the radius of the Earth.
\u2234 Acceleration due to gravity,<\/p>\n\n\n\n
a=Fm=GMR2From the above relation, we can see that acceleration produced in a body does not depend on the mass of the body. So, acceleration due to gravity is the same for all bodies.<\/p>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
Can you think of two particles which do not exert gravitational force on each other?<\/p>\n\n\n\n
Answer:<\/h4>\n\n\n\n
No. All practicals which have mass exert gravitational force on each other. Even massless particles experience the same gravitational force like other particles, because they do have relativistic mass.<\/p>\n\n\n\n
Question 5:<\/h4>\n\n\n\n
The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun? Or does it?<\/p>\n\n\n\n
Answer:<\/h4>\n\n\n\n
We know that the Earth-Moon system revolves around the Sun. The gravitational force of the Sun on the system provides the centripetal force its revolution. Therefore, the net force on the system is zero and the Moon does not experience any force from the Sun. This is the reason why the Moon revolves around the Earth and not around the Sun.<\/p>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
At noon, the sun and the earth pull the objects on the earth\u2019s surface in opposite directions. At midnight, the sun and the earth pull these objects in same direction. Is the weight of an object, as measured by a spring balance on the earth\u2019s surface, more at midnight as compared to its weight at noon?<\/p>\n\n\n\n
Answer:<\/h4>\n\n\n\n
No. Due to the revolution of the Earth around the Sun, the gravitational force of the Sun on the Earth system is almost zero. Hence, the body will not experience any force due to the Sun. Therefore, weight of the object will remain the same.<\/p>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g<\/em>. Who exerts the force needed to accelerate the earth with this acceleration g<\/em>?<\/p>\n\n\n\n The mutual gravitational force between the apple and the Earth is responsible for the acceleration produced in the apple falling from the tree. Although the Earth will experience the same force, it does not get attracted towards the apple because of its large mass. The insect feels that the Earth is falling towards the apple with an acceleration g<\/em> because of the the relative motion. Vea=velocity of earth w.r.t apple<\/p>\n\n\n\n vae=va-ve=-(ve-va)=-veaAs the insect is in the frame of apple so he sees the earth moving with a relative velocity<\/p>\n\n\n\n vea. vae.Both are opposite in direction.<\/p>\n\n\n\n Suppose the gravitational potential due to a small system is k<\/em>\/r<\/em>2<\/sup> at a distance r<\/em> from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?<\/p>\n\n\n\n The gravitational potential due to the system is given as<\/p>\n\n\n\n V=kr2. E=-dVdr\u21d2E=-ddrkr2=\u20132kr3\u21d2E=2kr3We can see that for this system,<\/p>\n\n\n\n E\u221d1r3This type of system is not possible because<\/p>\n\n\n\n Fgis always proportional to inverse of square of distance(experimental fact).<\/p>\n\n\n\n If there were negative masses, then this type of system is possible.<\/p>\n\n\n\n This system is a dipole of two masses, i.e., two masses, one positive and the other negative, separated by a small distance.<\/p>\n\n\n\n In this case, the gradational field due to the dipole is proportional to<\/p>\n\n\n\n 1r3.<\/p>\n\n\n\n The gravitational potential energy of a two-particle system is derived in this chapter as<\/p>\n\n\n\n U=Gm1m2r. Does it follow from this equation that the potential energy for<\/p>\n\n\n\n r=\u221emust be zero? Can we choose the potential energy for<\/p>\n\n\n\n r=\u221eto be 20 J and still use this formula? If no, what formula should be used to calculate the gravitational potential energy at separation r<\/em>?<\/p>\n\n\n\n The gravitational potential energy of a two-particle system is given by<\/p>\n\n\n\n U=-Gm1m2r. No, if we suppose that the potential energy for<\/p>\n\n\n\n r=\u221eis 20 J, then we need to modify the formula.<\/p>\n\n\n\n Now, potential energy of the two-particle system separated by a distance r<\/em> is given by<\/p>\n\n\n\n U(r,)=U(r)-U(\u221e)Given: U(\u221e)=20 J\u2234 U(r)=-Gm1m2r2-20This formula should be used to calculate the gravitational potential energy at separation r.<\/em><\/p>\n\n\n\n The weight of an object is more at the poles than at the equator. Is it beneficial to purchase goods at equator and sell them at the pole? Does it matter whether a spring balance is used or an equal-beam balance is used?<\/p>\n\n\n\n The weight of an object is more at the poles than that at the equator. In purchasing or selling goods, we measure the mass of the goods. The balance used to measure the mass is calibrated according to the place to give its correct reading. So, it is not beneficial to purchase goods at the equator and sell them at the poles. A beam balance measures the mass of an object, so it can be used here. For using a spring balance, we need to calibrate it according to the place to give the correct readings.<\/p>\n\n\n\n The weight of a body at the poles is greater than the weight at the equator. Is it the actual weight or the apparent weight we are talking about? Does your answer depend on whether only the earth\u2019s rotation is taken into account or the flattening of the earth at the poles is also taken into account?<\/p>\n\n\n\n The weight of a body at the poles is greater than that at the equator. Here, we are talking about the actual weight of the body at that particular place. If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what per cent?<\/p>\n\n\n\n If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by<\/p>\n\n\n\n g=GMR2. If the radius of the earth is decreased by 1%, then the new radius becomes<\/p>\n\n\n\n R\u2019=R-R100=99100R\u21d2R\u2019=0.99RNew acceleration due to gravity will be given by<\/p>\n\n\n\n g\u2019=GMR\u20192=GM(0.99R)2\u21d2g\u2019=1.02\u00d7GMR2=1.02gHence, the value of the acceleration due to gravity increases when the radius is decreased.<\/p>\n\n\n\n Percentage increase in the acceleration due to gravity is given by<\/p>\n\n\n\n g\u2019-gg\u00d7100 =0.02gg\u00d7100 =2%<\/p>\n\n\n\n A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on the earth? If yes, where will it land? If no, how can an astronaut make it land on the earth?<\/p>\n\n\n\n No, it will not land on the Earth. The nut will start revolving in the orbit of the satellite with the same orbital speed as that of the satellite due to inertia of motion. An astronaut can make it land on the Earth by projecting it with some velocity toward the Earth.<\/p>\n\n\n\n Is it necessary for the plane of the orbit of a satellite to pass through the centre of the earth?<\/p>\n\n\n\n According to Kepler first law of planetary motion all planets move in elliptical orbits with sun at one of its foci. It applies to any planet and its satellite as well.This implies that plane of the satellite has to pass through the centre of planet (earth).<\/p>\n\n\n\n Consider earth satellites in circular orbits. A geostationary satellite must be at a height of about 36000 km from the earth\u2019s surface. Will any satellite moving at this height be a geostationary satellite? Will any satellite moving at this height have a time period of 24 hours?<\/p>\n\n\n\n T=(gR2T24\u03c02)13-RT=4\u03c02(h+R)3gR2=4\u00d73.142\u00d7(36000+6400)3\u00d71099.8\u00d7(6400\u00d7103)2=24.097 HrWhich implies that it is a geostationary sattelite with time period=24 Hrs.<\/p>\n\n\n\n No part of India is situated on the equator. Is it possible to have a geostationary satellite which always remains over New Delhi?<\/p>\n\n\n\n No. All geostationary orbits are concentric with the equator of the Earth.<\/p>\n\n\n\n As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he\/she not feel weightless as a satellite passenger does?<\/p>\n\n\n\n A person living in a house at the equator will not feel weightlessness because he is not in a free fall motion. Satellites are in free fall motion under the gravitational pull of the earth, but, due to the curved surface of the Earth, they move in a circular path. The gravitational force on the satellite due to the Sun provides the centripetal force for its revolution. Therefore, net force on the satellite is zero and, thus, a person feels weightless in a satellite orbiting the earth.<\/p>\n\n\n\n Two satellites going in equatorial plane have almost same radii. As seen from the earth one moves from east one to west and the other from west to east. Will they have the same time period as seen from the earth? If not which one will have less time period?<\/p>\n\n\n\n No, both satellites will have different time periods as seen from the Earth. The satellite moving opposite (east to west) to the rotational direction of the Earth will have less time period, because its relative speed with respect to the Earth is more.<\/p>\n\n\n\n A spacecraft consumes more fuel in going from the earth to the moon than it takes for a return trip. Comment on this statement.<\/p>\n\n\n\n Yes, a spacecraft consumes more fuel in going from the Earth to the Moon than it takes for the return trip. In going from the Earth to the Moon, the spacecraft has to overcome the gravitational pull of the earth. So, more fuel is consumed in going from the Earth to Moon. However, in the return trip, this gravitation pull helps the spacecraft to come back to the Earth.<\/p>\n\n\n\n The acceleration of moon with respect to earth is 0\u22c50027 m s\u22122<\/sup> and the acceleration of an apple falling on earth\u2019 surface is about 10 m s\u22122<\/sup>. Assume that the radius of the moon is one fourth of the earth\u2019s radius. If the moon is stopped for an instant and then released, it will fall towards the earth. The initial acceleration of the moon towards the earth will be (b) 0\u22c50027 m s\u22122<\/sup><\/p>\n\n\n\n We know that the distance of the Moon from the Earth is about 60 times the radius of the earth. So, acceleration due to gravity at that distance is 0.0027 m\/s2<\/sup>. When the Moon is stopped for an instant and then released, it will fall towards the Earth with an initial acceleration of 0.0027 m\/s2<\/sup>.<\/p>\n\n\n\n The acceleration of the moon just before it strikes the earth in the previous question is (c) 6\u22c54 m s\u22122<\/sup><\/p>\n\n\n\n According to the previous question, we have: Rm=Re4=64000004=1600000 mSo, when the Moon is just about to hit the surface of the Earth, its centre of mass is at a distance of (Re<\/sub> + Rm<\/sub><\/em>) from the centre of the Earth.<\/p>\n\n\n\n Acceleration of the Moon just before hitting the surface of the earth is given by<\/p>\n\n\n\n g\u2019=GM(Re+Rm)2=GMRe2(1+RmRe)2\u21d2g\u2019=g(1+RmRe)2=10(1+14)2=10\u00d71625\u21d2g\u2019=6.4 m\/s2<\/p>\n\n\n\n Suppose, the acceleration due to gravity at the earth\u2019s surface is 10 m s\u22122<\/sup> and at the surface of Mars it is 4\u22c50 m s\u22122<\/sup>. 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 (11-Q1) best represents the weight (net gravitational force) of the passenger as a function of time? (c) C<\/em><\/p>\n\n\n\n At one point between the Earth and Mars, the gravitational field intensity is zero. So, at that point, the weight of the passenger is zero. The curve C indicates that the weight of the passenger is zero at a point between the Earth and Mars.<\/p>\n\n\n\n Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W<\/em> on the earth will weight (d) 21\/3<\/sup> W<\/em> at the planet<\/p>\n\n\n\n The weight of the object on the Earth is<\/p>\n\n\n\n W=mGMeRe2. Mp=2MeIf<\/p>\n\n\n\n \u03c1is the average density of the planet then Wp=mGMpRp2=m2GMe223Re2\u21d2Wp=213\u00d7mGMeRe2\u21d2Wp=213\u00d7W<\/p>\n\n\n\n If the acceleration due to gravity at the surface of the earth is g<\/em>, the work done in slowly lifting a body of mass m<\/em> from the earth\u2019s surface to a height R<\/em> equal to the radius of the earth is 12mgR(b)<\/p>\n\n\n\n 2mgR(c)<\/p>\n\n\n\n mgR(d)<\/p>\n\n\n\n 14mgR.<\/p>\n\n\n\n (a)<\/p>\n\n\n\n 12mgRWork done =<\/p>\n\n\n\n -(final potential energy<\/p>\n\n\n\n -initial potential energy)<\/p>\n\n\n\n \u21d2W=-GMm2R-GMmR\u21d2W=12GMmR=12mR\u00d7GMR2\u21d2W=12mRg \u2235 g=GMR2<\/p>\n\n\n\n A person brings a mass of 1 kg from infinity to a point A<\/em>. Initially the mass was at rest but it moves at a speed of 2 m s \u22121<\/sup> as it reaches A<\/em>. The work done by the person on the mass is \u22123 J. The potential at A<\/em> is The work done by the person is equal to the kinetic energy and the potential energy of the mass of 1 kg at point A.<\/p>\n\n\n\n Let V<\/em>A<\/sub> be the potential at point A.<\/p>\n\n\n\n Now, W=12mv2+P.E.A\u21d2W=12mv2+VA\u00d7m\u21d2-3=12\u00d71\u00d7(2)2+VA\u00d71\u21d2VA=-5 J\/Kg<\/p>\n\n\n\n Let V<\/em> and E<\/em> be the gravitational potential and gravitational field at a distance r<\/em> from the centre of a uniform spherical shell. Consider the following two statements: (c) B<\/em> is correct but A<\/em> is wrong.<\/p>\n\n\n\n The plot of E<\/em> against r<\/em> is discontinuous as gravitational field inside the spherical shell is zero (r < R<\/em>). The plot of V<\/em> against r<\/em> is a continuous curve for a uniform spherical shell.<\/p>\n\n\n\n Let V<\/em> and E<\/em> represent the gravitational potential and field at a distance r<\/em> from the centre of a uniform solid sphere. Consider the two statements: (d) Both A<\/em> and B<\/em> are wrong.<\/p>\n\n\n\n Both the plots (i.e., V<\/em> against r<\/em> and E<\/em> against r<\/em>) are continuous curves for a uniform solid sphere.<\/p>\n\n\n\n Take the effect of bulging of earth and its rotation in account. Consider the following statements: (b) A<\/em> is correct but B<\/em> is wrong.<\/p>\n\n\n\n We know that the value of acceleration due to gravity decreases when we go up from the surface of the Earth. If we take the into account the effect of bulging of the Earth due to its rotation, we can say that acceleration due to gravity is maximum at the poles and minimum at the equator<\/p>\n\n\n\n So, there are points above both the poles where the value of g<\/em> is equal to its value at the equator.<\/p>\n\n\n\n Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them.<\/p>\n\n\n\n The gravitational force of attraction between the balls is given as by<\/p>\n\n\n\n F=Gm1m2r2Given: m1=m2=10 kg and r=10 cm=0.10 m<\/p>\n\n\n\n \u2234 F=6.67\u00d710-11\u00d710\u00d7100.12 \u21d2 F=6.67\u00d710-7 N<\/p>\n\n\n\n Four particles having masses m<\/em>, 2m<\/em>, 3m<\/em> and 4m<\/em> are placed at the four corners of a square of edge a<\/em>. Find the gravitational force acting on a particle of mass m<\/em> placed at the centre.<\/p>\n\n\n\n Force due to the particle at A, F\u2192OA=G\u00d7m\u00d7mOA2Let OA =r\u2234 F\u2192OA=G\u00d7m\u00d7mr2 Here, r=a22+a22=a2Force due to the particle at B, F\u2192OB=G\u00d7m\u00d72mr2Force due to the particle at C, F\u2192OC=G\u00d7m\u00d73mr2Force due to the particle at D, F\u2192OD=G\u00d7m\u00d74mr2Now, resultant force=F\u2192OA+F\u2192OB+F\u2192OC+F\u2192OD=2Gmma2-i\u21922+j\u21922+4Gmma2i\u21922+j\u21922=6Gmma2i\u21922-j\u21922+8Gmma2-i\u21922\u2013j\u21922\u2234 F=44Gm2a2 j\u23dc<\/p>\n\n\n\n The time period of an earth-satellite in circular orbit is independent of (a) the mass of the satellite<\/p>\n\n\n\n The time period of an earth-satellite in circular orbit is independent of the mass of the satellite, but depends on the radius of the orbit.<\/p>\n\n\n\n The magnitude of gravitational potential energy of the moon-earth system is U<\/em> with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K<\/em>. (b) U<\/em> > K<\/em><\/p>\n\n\n\n For a system to be bound , total energy of the system should be negative.As we know that kinetic energy can never be negative. Figure (11-Q2) shows the elliptical path of a planet about the sun. The two shaded parts have equal area. If t<\/em>1<\/sub> and t<\/em>2<\/sub> be the time taken by the planet to go from a<\/em> to b<\/em> and from c<\/em> to d<\/em> respectively, (b) t<\/em>1<\/sub> = t<\/em>2<\/sub><\/p>\n\n\n\n Kepler\u2019s second law states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time, i.e., areal velocity of a planet about the Sun is constant.<\/p>\n\n\n\n The given areas swept by the planet are equal, so t<\/em>1<\/sub> = t<\/em>2<\/sub><\/p>\n\n\n\n A person sitting in a chair in a satellite feels weightless because (c) the normal force is zero<\/p>\n\n\n\n The gravitational pull on the satellite is used up in providing the necessary centripetal force required for its revolution around the earth. This means that there is no net force on the person sitting in a chair in the satellite. So, the normal reaction of the chair on the person is zero and he will feel weightless.<\/p>\n\n\n\n A body is suspended from a spring balance kept in a satellite. The reading of the balance is W<\/em>1<\/sub>when the satellite goes in an orbit of radius R and is W<\/em>2<\/sub> when it goes in an orbit of radius 2 \u2212R<\/em>. W1\u2260W2.<\/p>\n\n\n\n (a) W<\/em>1<\/sub> = W<\/em>2<\/sub><\/p>\n\n\n\n The gravitational pull on the satellite in both cases is used up in providing the necessary centripetal force required for its revolution around the earth. This means that there is no net force acting on the body which has been suspended from a spring balance in the satellite. So, the readings of the spring balance in both the cases are the same and is equal to zero.<\/p>\n\n\n\n The kinetic energy needed to project a body of mass m<\/em> from the earth\u2019 surface to infinity is 14mgR(b)<\/p>\n\n\n\n 12mgR(c)<\/p>\n\n\n\n mgR(d)<\/p>\n\n\n\n 2 mgR.<\/p>\n\n\n\n (c)<\/p>\n\n\n\n mgRThe kinetic energy needed to project a body of mass m<\/em> from the Earth\u2019s surface to infinity is equal to the negative of the change in potential energy of the body.<\/p>\n\n\n\n i.e., kinetic energy =<\/p>\n\n\n\n -(final potential energy<\/p>\n\n\n\n -initial potential energy)<\/p>\n\n\n\n \u21d2K=-GMmR\u2019-GMmR\u21d2K=-GMm\u221e-GMmR=GMmR\u21d2K=mR\u00d7GMR2\u21d2K=mRg \u2234 g=GMR2<\/p>\n\n\n\n A particle is kept at rest at a distance R<\/em> (earth\u2019s radius) above the earth\u2019s surface. The minimum speed with which it should be projected so that it does not return is GM4R(b)<\/p>\n\n\n\n GM2R(c)<\/p>\n\n\n\n GMR(d)<\/p>\n\n\n\n 2GMR<\/p>\n\n\n\n (c)<\/p>\n\n\n\n GMRPotential energy of the particle at a distance R from the surface of the Earth is<\/p>\n\n\n\n P.E.i=GMm(R+R)=12GMmR. Let the particle be projected with speed v<\/em> so that it just escapes the gravitational pull of the earth. -[change in the potential energy of the body]<\/p>\n\n\n\n Now, kinetic energy of the body =<\/p>\n\n\n\n -[final potential energy<\/p>\n\n\n\n -initial potential energy]<\/p>\n\n\n\n \u21d212mv2=-GMm\u221e-GMm2R\u21d2v=GMmR<\/p>\n\n\n\n A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is<\/p>\n\n\n\n \u03bde. Its speed with respect to the satellite \u03bde(b) will be more than<\/p>\n\n\n\n \u03bde(c) will be equal to<\/p>\n\n\n\n \u03bde(d) will depend on the direction of projection.<\/p>\n\n\n\n (d) will depend on the direction of projection.For example a body projected vertically requires less escape velocity than a body projected at an angle with the vertical.<\/p>\n\n\n\n Let V<\/em> and E<\/em> denote the gravitational potential and gravitational field at a point. It is possible to have V=0 and E=0(b)<\/p>\n\n\n\n V=0 and E\u22600(c)<\/p>\n\n\n\n V\u22600 and E=0(d)<\/p>\n\n\n\n V\u22600 and E\u22600.<\/p>\n\n\n\n (a)<\/p>\n\n\n\n V=0 and E=0(b)<\/p>\n\n\n\n V=0 and E\u22600(c)<\/p>\n\n\n\n V\u22600 and E=0(d)<\/p>\n\n\n\n V\u22600 and E\u22600.<\/p>\n\n\n\n All the given conditions for gravitational potential and gravitational field at a point are possible.<\/p>\n\n\n\n Inside a uniform spherical shell (b) the gravitational field is zero Inside a uniform spherical shell, the gravitational field is the same everywhere and is equal to zero. The gravitational potential has a constant value inside a uniform spherical shell.<\/p>\n\n\n\n A uniform spherical shell gradually shrinks maintaining its shape. The gravitational potential at the centre (b) decreases Consider a planet moving in an elliptical orbit round the sun. The work done on the planet by the gravitational force of the sun (b) is zero in some parts of the orbit When a planet is moving in an elliptical orbit, at some point, the line joining the centre of the Sun and the planet is perpendicular to the velocity of the planet. For that instant, work done by the gravitational force on the planet becomes zero. As there is no net increase in the speed of the planet after one complete revolution about the Sun, the work done by the gravitational force on the planet in one complete revolution is zero. Two satellites A<\/em> and B<\/em> move round the earth in the same orbit. The mass of B<\/em> is twice the mass of Answer:<\/h4>\n\n\n\n
let
vae<\/sub>=velocity of apple w.r.t earth<\/p>\n\n\n\n
Any other observer on earth will see the apple moving towards earth with velocity<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Gravitational field due to the system:<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
This relation does not tell that the gravitational potential energy is zero at infinity. For our convenience, we choose the potential energies of the two particles to be zero when the separation between them is infinity.<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Yes. If the rotation of the Earth is taken into account, then we are discussing the apparent weight of the body.<\/p>\n\n\n\nPage No 224:<\/h4>\n\n\n\n
Question 12:<\/h4>\n\n\n\n
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Here, M<\/em> is the mass of Earth; R<\/em> is the radius of the Earth and G<\/em> is universal gravitational constant.<\/p>\n\n\n\nQuestion 13:<\/h4>\n\n\n\n
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Question 1:<\/h4>\n\n\n\n
(a) 10 m s\u22122<\/sup>
(b) 0\u22c50027 m s\u22122<\/sup>
(c) 6\u22c54 m s\u22122<\/sup>
(d) 5\u22c50 m s\u22122<\/sup>.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
(a) 10 m s\u22122<\/sup>
(b) 0\u22c50027 m s\u22122<\/sup>
(c) 6\u22c54 m s\u22122<\/sup>
(d) 5\u22c50 m s\u22122<\/sup><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Radius of the moon,<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(a) A<\/em>
(b) B<\/em>
(c) C<\/em>
(d) D<\/em>.
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
(a) W<\/em>
(b) 2 W<\/em>
(c) W<\/em>\/2
(d) 21\/3<\/sup> W<\/em> at the planet.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Here, m<\/em> is the actual mass of the object; Me <\/sub><\/em>is the mass of the earth and Re<\/sub><\/em> is the radius of the earth.
Let Rp<\/sub> be the radius of the planet.
Mass of the planet,<\/p>\n\n\n\n
43\u03c0Rp3\u00d7\u03c1=2\u00d743\u03c0Re3\u00d7\u03c1\u21d2Rp=213ReNow, weight of the body on the planet is given by<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(a) \u22123 J kg\u22121<\/sup>
(b) \u22122 J kg\u22121<\/sup>
(c) \u22125 J kg\u22124<\/sup>
(d) none of these.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
(A) The plot of V<\/em> against r<\/em> is discontinuous.
(B) The plot of E<\/em> against r<\/em> is discontinuous.
(a) Both A<\/em> and B<\/em> are correct.
(b) A<\/em> is correct but B<\/em> is wrong.
(c) B<\/em> is correct but A<\/em> is wrong.
(d) Both A<\/em> and B<\/em> are wrong.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/7Q12.png\")
<\/figure>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
(A) the plot of V<\/em> against r<\/em> is discontinuous.
(B) The plot of E<\/em> against r<\/em> is discontinuous.
(a) Both A<\/em> and B<\/em> are correct.
(b) A<\/em> is correct but B<\/em> is wrong.
(c) B<\/em> is correct but A<\/em> is wrong.
(d) Both A<\/em> and B<\/em> are wrong.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
(A) There are points outside the earth where the value of g<\/em> is equal to its value at the equator.
(B) There are points outside the earth where the value of g<\/em> is equal to its value at the poles.
(a) Both A<\/em> and B<\/em> are correct.
(b) A<\/em> is correct but B<\/em> is wrong.
(c) B<\/em> is correct but A<\/em> is wrong.
(d) Both A<\/em> and B<\/em> are wrong.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Page No 225:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/2_exercises2.png\")
<\/figure>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
(a) the mass of the satellite
(b) radius of the orbit
(c) none of them
(d) both of them.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/10209.png\")
<\/figure>\n\n\n\nQuestion 11:<\/h4>\n\n\n\n
(a) U<\/em> < K<\/em>
(b) U<\/em> > K<\/em>
(c) U<\/em> = K<\/em>.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
E<0 & K>0 & E=K+U
it gives that U>K.<\/p>\n\n\n\nQuestion 12:<\/h4>\n\n\n\n
(a) t<\/em>1<\/sub> < t<\/em>2<\/sub>
(b) t<\/em>1<\/sub> = t<\/em>2<\/sub>
(c) t<\/em>1<\/sub> > t<\/em>2<\/sub>
(d) insufficient information to deduce the relation between t<\/em>1<\/sub> and t<\/em>2<\/sub>.
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 13:<\/h4>\n\n\n\n
(a) the earth does not attract the objects in a satellite
(b) the normal force by the chair on the person balances the earth\u2019s attraction
(c) the normal force is zero
(d) the person in satellite is not accelerated.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 14:<\/h4>\n\n\n\n
(a) W<\/em>1<\/sub> = W<\/em>2<\/sub>
(b) W<\/em>1<\/sub> < W2<\/sub>
(c) W<\/em>1<\/sub> > W<\/em>2<\/sub>
(d)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 15:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 16:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Here, M<\/em> is the mass of the earth; R<\/em> is the radius of the earth and m<\/em> is the mass of the body.<\/p>\n\n\n\n
So, kinetic energy of the body =<\/p>\n\n\n\nQuestion 17:<\/h4>\n\n\n\n
(a) will be less than<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
(a) the gravitational potential is zero
(b) the gravitational field is zero
(c) the gravitational potential is same everywhere
(d) the gravitational field is same everywhere<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) the gravitational potential is same everywhere
(d) the gravitational field is same everywhere<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(a) increases
(b) decreases
(c) remains constant
(d) oscillates.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
AS it is maintaining its shape so its mass will remain constant which implies that if it shrinks then its volume decreases and to keep the mass constant ,its density increases.Also the gravitational potential at the centre of a uniform spherical shell is inversely proportional to the radius of the shell with a negative sign. When a uniform spherical shell gradually shrinks, the gravitational potential at the centre decreases (because of the negative sign in the formula of potential).<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(a) is zero in any small part of the orbit
(b) is zero in some parts of the orbit
(c) is zero in one complete revolution
(d) is zero in no part of the motion.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) is zero in one complete revolution<\/p>\n\n\n\n
Note:For elliptical orbits angle between force ans velocity is always 90 so there the work done is zero in any small part of the orbit.<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n