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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
Correct the following. (a) Force of attraction increases when an object is raised from earth’s surface (b) Value of g remains the same on all regions of earth. |
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Answer» (a) Force of attraction decreases when an object is raised from earth’s surface. (b) Value of g is different on all regions of earth. |
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| 852. |
Which is greater out of the following : (i) The attraction of earth of 5 kg of copper? (ii) The attraction of 5 kg copper for earth? |
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Answer» Same. Both are same. BOTH ARE SAME... |
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| 853. |
A satellite revolves around the earth along a circular path. If the mass of the satellite is 1000 kg and its distance from the center of the earth is 20000 km, find the magnitude of the earth’s gravitational force acting on the satellite. |
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Answer» Answer is 1000.5 |
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| 854. |
(a) What is mean by free fall? (b) Weight of an object in free fall is zero? Why. (c) If a bottle having a hole at the bottom filled with water falling down freely water will not flow out. Why? |
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Answer» (a) Falling of a body towards earth due to force of gravity is said to be free fall. (b) In free fall the object cannot be able to give reaction force. Also, the gravitational force is utilized to give acceleration to the object. (c) During free fall, the acceleration of water and bottle remains the same. So no reaction force feel in water. |
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| 855. |
Complete the table |
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| 856. |
Where does a body weight more at the pole or at the equator? |
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Answer» It weighs more at the pole. |
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| 857. |
Find the acceleration due to gravity at a distance of 20000 km from the center of the earth. |
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Answer» Answer is 1.0 m/s2 |
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| 858. |
Fill in the blanks.(a) Acceleration due to gravity does not affect the ……………. of the object (b) 1 kg wt = ………….. N |
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Answer» (a) Mass (b) 9.8 |
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| 859. |
How the mass and distance between objects affect gravitational force? |
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Answer» Among the two bodies if the mass of one body is doubled gravitational force becomes two times. When the mass of both the bodies are doubled. Gravitational force becomes four times. When the distance increases gravitational force decreases. Gravitational force varies in reciprocal of the square of the distance between the bodies. |
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| 860. |
A body weight 1400 gram weight on the surface of earth. How will it weight on the surface of a planet whose mass is `(2)/(7)` and radius is `(1)/(3)` that of the earth ?A. 0.45 kg wtB. 0.9 kg wtC. 1.8 kg wtD. 3.6 kg wt |
| Answer» Correct Answer - D | |
| 861. |
What is the weight of a body of mass 100 kg at the south pole? (g = 9.832 m/s2) |
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Answer» Answer is 983.2 N (downward) |
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| 862. |
Mass and weight of a body is determined at the pole and at the equator (a) Is there any difference in the mass? (b) Is there any change in the weight? (c) Justify the answer |
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Answer» (a) No change in mass (b) Change will occur (c) Mass is the quantity of matter present in a matter. Value of g is greater at the poles and lesser at the equator. So weight varies. Value of g does not affect the mass |
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| 863. |
A person moves from pole to equator, then his weight willA. IncreaseB. DecreaseC. Remain sameD. First increase and then decrease |
| Answer» Correct Answer - B | |
| 864. |
What is the weight of a body of mass 20 kg at the equator? (g = 9.78 m/s2) |
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Answer» Answer is 195.6 N (downward) |
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| 865. |
A body is released from the top of a tower of height 50 m. Find the velocity with which the body hits the ground, (g = 9.8 m/s2) |
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Answer» Answer is 31.3 m/s (downward) |
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| 866. |
A body is released from the top of a building of height 19.6 m. Find the velocity with which the body hits the ground. |
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Answer» Data: h = 19.6 m, u = 0 m/s, g = 9.8 m/s2, s = 19.6 m, v = ? v2 = u2 + 2 gs . = 2 gs …..(as u = 0 m/s) = 2 × 9.8 m/s2 × 19.6 m = (19.6 m/s)2 ∴ v = 19.6 m/s (downward velocity) The velocity with which the body hits the i ground = 19.6 m/s (downward). |
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| 867. |
A uniform ring of mass m and radius 3a is kept above a sphere of mass M and radius 3a at a distance of 4a (as shown in figure) such that line joining the centres of ring and sphere is perpendicular to the plane of the ring. Find the force of gravitational attraction between the ring and the sphere. |
| Answer» Correct Answer - `(4GMm)/(125a^(2))` | |
| 868. |
The ratio of the escape speed from two planets is 3 : 4 and the ratio of their masses is 9 : 16. What is the ratio of their radii ? |
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Answer» Correct Answer - `1 : 1` Hint : Take `(v_(1))/(v_(2))=sqrt((M_(1))/(M_(2))xx(R_(2))/(R_(1)))` |
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| 869. |
What will be the escape speed from earth if the mass of earth is doubled keeping the same size ? |
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Answer» Correct Answer - 15.83 km/s Hint Use `v=sqrt((2GM)/(R ))` |
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| 870. |
A fixed spherical shell of inner radius R and outer radius 2R has mass M, A particle of mass m is held at rest at a distance 2R from the surface of the body. If the particle is released from that point, with what speed will it hit surface of the spherical body ? |
| Answer» Correct Answer - `sqrt((GM)/(2R))` | |
| 871. |
With what minimum velocity v should a particle be thrown horizontally from a height h above the surface of the earth of mass M so that it does not fall back on the earth ? Radius of earth is R ? |
| Answer» `sqrt((2GMR)/((2R+h)(R+h)))` | |
| 872. |
An electron of mass `9.1xx10^(-31)` kg is at a distance of `10 A^(@)` from a. porton of mass `1.67xx10^(-27)` kg. calculate the gravitational force of attraction between them. |
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Answer» Here, `m_(1)=9.1xx10_(-31) kg, d=10A^(@)=10^(-9)m` `M_(2)=1.67xx10^(27)kg, F=?` `F=(Gm_(1)m_(2))/d^(2)=(6.67xx10_(-11)xx9.1xx10^(-31)xx1.67xx10^(-27))/(10^(-9))^(2)=1.01xx10^(-49)N` |
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| 873. |
The escape velocity of a body form the earth depends on (i) the mass of the body. (ii) the location from where it is projected. (iii) the direction of projection. (iv) the height of the location form where the body is launched.A. (i) and (ii)B. (ii) and (iv)C. (i) and (iii)D. (iii) and (iv) |
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Answer» Correct Answer - B The escape velocity is indepent of mass of the body and the direction of projection. It depeds upon the gravitational potential at the point from where the body is lauched. Since this potential depends slightly on the latitude and height of the point, the escape velocity depends slightely on these factors. |
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| 874. |
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun. |
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Answer» No No Yes No No Yes Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic, and potential energy varies from point to point in the orbit. |
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| 875. |
The gravitational force between two bodies is `6.67xx10^(-7) N` when the distance between their centres is `10m`. If the mass of first body is `800 kg`, then the mass of second body isA. `1000 kg`B. `1250 kg`C. `1500 kg`D. `2000 kg` |
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Answer» Correct Answer - B `F_(g)=(Gm_(1)m_(2))/(R^(2))rArr m_(2)=(F_(g)xxR^(2))/(Gm_(1))` |
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| 876. |
The energy required to move an earth satellites of mass m from a circular orbit of radius 2 R to a radius 3 R is `" "` (R is radius of the earth)A. `(GMm)/(R)`B. `(GMm)/(2R)`C. `(GMm)/(12R)`D. `(GMm)/(4R)` |
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Answer» Correct Answer - c `E(2)-E_(1)=-(GMm)/(2r_(2))+(GMm)/(2r_(1))` `=-(GMm)/(6R)+(GMm)/(4R)` `=(GMm)/(2R)((1)/(2)-(1)/(3))=(GMm)/(12R)`. |
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| 877. |
If the weight of a body on the earth is 6 N, what will it be on the moon ? |
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Answer» Weight of the body on the surface of moon will be 1N approx. as the value of g on the surface of moon is one-sixth that of the earth |
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| 878. |
The mass of a body on the surface of earth is 12. if acceleration due to gravity on moon is `(1)/(6)` of acceleration due to gravity on earth, then its mass on moon will beA. `2kgf`B. 72kgC. 12kgD. zero |
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Answer» Correct Answer - C Mass of body`=12kg`,as it is not affected by gravity. |
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| 879. |
The mass of the Earth is `6 xx 10^(24)` kg and its radius is 6400 km. Find the acceleration due to gravity on the surface of the Earth. |
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Answer» Given Mass of the Earth, `M = 6 xx 10^(24) kg` Radius of the Earth, `R = 6400 km = 64 xx 10^(5) m` Universal gravitational constant, `G = 6.67 xx 10^(-11) N m^(2) kg^(-2)` Acceleration due to gravity on the surface of the Earth, `g = (GM)/(R^(2))` `" "g = (GM)/(r^(2))=(6.67 xx 10^(-11) xx 6 xx 20^(24))/((64 xx 10^(5)))=9.77 m s^(-2)` |
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| 880. |
When a stone is thrown up, it reach it reaches a certain height and then stars falling down. Why ? |
| Answer» When a stone is thrown up,its velocity goes on decreasing at a constant rate of `9.8m//s^(2)`. At a certain height, its velocity becomes zero. So it can not rise further. From this maximum height, the stone falls down with constant acceleration due to `gravity=9.8m//s^(2)`. | |
| 881. |
Lifting a small stone to a certain height and then dropping it downwards.What do you observe? |
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Answer» The stone falls down |
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| 882. |
When a stone of mass m is falling on the earth of mass M; find the acceleration of earth if any? |
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Answer» Force exerted by falling stone on earth, F = mg Acceleration of earth = F/M = mg/M |
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| 883. |
Lifting a small stone to a certain height and then dropping it downwards.What could be the reason for the falling of the stone? |
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Answer» Earth’s attraction is the reason |
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| 884. |
Lifting a small stone to a certain height and then dropping it downwards.What change takes place in the speed of the stone as it is thrown up? |
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Answer» Speed decreases |
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| 885. |
Lifting a small stone to a certain height and then dropping it downwards.What about the speed when it falls down? |
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Answer» Speed increases |
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| 886. |
Lifting a small stone to a certain height and then dropping it downwards.Did you apply any force on the stone to bring it down? |
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Answer» Answer is No |
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| 887. |
Lifting a small stone to a certain height and then dropping it downwards.From where did the stone get the force for the acceleration? |
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Answer» The force required for acceleration got from earth’s attraction |
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| 888. |
The stone tied to a thread is suspending from a spring balance.What do you observe? |
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Answer» Spring is stretched downwards |
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| 889. |
The stone tied to a thread is suspending from a spring balance.The spring stretched down when the stone was suspended from it. Why? |
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Answer» Earth’s attraction is the reason The earth attracts all objects towards its centre. This force of attraction is the force of gravity. Conclusion:
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| 890. |
Write down instances where the force of gravity is felt. |
Answer»
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| 891. |
What is the value of gravitational constant G (i) on the earth, and (ii) on the moon ? |
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Answer» Value of gravitational constant G on the earth and the moon is = 6.67 x 10-11 Nm2 /kg2 Note that the value of G always remains constant irrespective of the location. |
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| 892. |
Particles of masses 2M, m and M are respectively at points A, B and C with `sqrt((GM)/(R))(sqrt(2)+1)` is much-much smaller than M and at time `t=0`, they are all at rest as given in figure At subsequent time before any collision takes place.A. `m` will remain at restB. `m` will move towards MC. `m` will move towards 2MD. `m` will have oscillatory motion |
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Answer» Correct Answer - C Fore on B due to A `=F_(BA)=(G(2Mm))/((AB)^(2))` towards BA Force on B due to C `= F_(BC)=(GMm)/((BC)^(2))` towards BC As, `(BC)=2AB` `rArr F_(BC)=(GMm)/((2AB)^(2))=(GMm)/(4(AB)^(2))lt F_(BA)` Hence, `m` will move towards BA (i.e., 2M) |
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| 893. |
Two satellite of same mass are launched in the same orbit of radius `r` around the earth so as to rotate opposite to each other. If they collide inelastically and stick together as wreckage, the total energy of the system just after collision isA. `-(2GMm)/(r)`B. `-(GMm)/(r)`C. `(GMm)/(2r)`D. zero |
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Answer» Correct Answer - A If they collide inelastically and stick together as wrekage, then KE = 0, only PE is present `:. E=U=2(-(GMm)/(r))=-(2GMm)/(r)` . |
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| 894. |
Choose the correct alternative: 1. If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of Its kinetic/ potential energy. 2. The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence. |
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Answer» 1. Kinetic energy: The potential energy of a satellite rotating in its orbit is zero. The total energy of a system is the sum of its kinetic energy (+ve) and potential energy. Since the earth satellite system is a bound system. The satellite has a negative total energy. So, the energy of the satellite is the negative of its kinetic energy. 2. Less: An orbiting satellite has more energy than a stationary object at the same height. This additional energy is provided by the orbit. It requires lesser energy to make it move out of the earth’s influence than a stationary object. |
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| 895. |
Does the escape speed of a body from the Earth depend on(a) the mass of the body(b) the location from where it is projected(c) the direction of projection(d) the height of the location from where the body is launched?Explain your answer. |
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Answer» (a) The escape speed of a body is independant of the mass of a body (b) The escape speed of a body depends upon the location (c) The escape speed of a body is independent of the direction of projection (d) The escape speed of a body depends upon the height of the location from where the body is projected, because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also. |
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| 896. |
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant(a) linear speed(b) angular speed(c) angular momentum(d) Kinetic energy(e) potential energy(f) total energy throughout its orbit?Neglect any mass loss of the comet when it comes very close to the sun. |
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Answer» Angular momentum and total energy do not vary throughout the orbit, whereas rest, all quantities vary in the orbit. |
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| 897. |
Does the escape speed of a body from the earth depend on? 1. the mass of the body, 2. the location from where it is projected, 3. the direction of projection, 4. the height of the location from where the body is launched? |
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Answer» Escape velocity of a body from earth is given by v = √(2gR). g = Acceleration due to Gravity. R = Radius of earth. 1. No. Escape velocity is independent of mass of the object. 2. No. Assuming the value of g is the same in all the locations, it is independent of location of projection. 3. No. The direction of projection is also immaterial as long as it is projected with the velocity above ground. 4. Yes. It depends on the gravitational potential energy which depends on the height. Since g changes with height. The escape velocity also depends on height. |
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| 898. |
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant? 1. linear speed, 2. angular speed, 3. angular momentum, 4. kinetic energy, 5. potential energy, 6. total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun. |
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Answer» 1. No, linear speed varies from point to point in the orbit. 2. No, angular speed varies from point to point in the orbit. 3. Yes, angular momentum is conserved at any point 4. No, kinetic energy varies as the speed varies. 5. No, potential energy varies as distance, from Sun varies. 6. Yes, the total energy of the system is always constant. |
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| 899. |
Haley’s Comet is going around the Sun in a highly elliptical orbit with a period of 76 y. It was closest to the sun in the year 1987 (I was 13 year old then and heard a lot about it n radio). In which year of `21^(st)` century do you expect it to have least kinetic energy? |
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Answer» Correct Answer - `4xx10^(16)m^(2)s^(-s)` |
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| 900. |
Two identical spheres are placed in contact with each other. The force of gravitation between the spheres will be proportional to ( R = radius of each sphere)A. `R`B. `R^(2)`C. `R^(4)`D. None of these |
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Answer» Correct Answer - C |
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