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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.
| 1501. |
Statement-1 :The universal gravitational constant is same as acceleration due to gravity. Statement-2 :Gravitional constant and acceleration due to gravity have different dimensional formulaA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - D |
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| 1502. |
The orbit of geostationary satellite is circular, the time period of satellite depeds on (i) mass of the satellite, (ii) mass of earth, (iii) readius of the orbit and (iv) height of the satellite from the surface of the earthA. (i) onlyB. (i) and (ii)C. (i),(ii) and (iii)D. (ii),(iii) and (iv) |
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Answer» Correct Answer - D (d) Orbital velocity ,` v_(0)=sqrt((GM_(E))/(R_(E)+h))` Time period, `T=(2pi(R_(E)+h))/(v_(0))=(2pi(R_(E)+h)^(3//2))/((GM_(E))^(1//2))` Thus, the time period of satellite is independent of mass of satellite but depends on mass of the earth, radius of the orbit `(R_(E)+h)`, height of the satellite from the surface of the earth. |
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| 1503. |
The gravitational potential change uniformly from `-20 J//kg` to `-40J//kg` as one moves along `X`-axis from `x=-1m` to `x=+1m`. Mark the correct statement (s) about gravitational field intensity of origin.A. The gravitational field intensity at `x=0` must be equal to `10N//kg`.B. The gravitational field intensity at `x=0` may be equal to `10N//kg`.C. The gravitational field intensity at `x=0` may be greater than `10N//kg`.D. The gravitational field intensity at `x=0` must not be less than `10 N//kg`. |
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Answer» Correct Answer - B::C::D In a head on the collision between two particles, the kinetic energy becomes minimum and potential energy becomes maximum and potential energy becomes maximum at the instant when and energy are conserved at energy instant. Let `m` and `u` be the mass and initial velocity of the first particles, `2m` be the mass of second particle and `v` be the common velocity. then, `1/2m u^(2)=3J,m u=(m+2m)v` or `v=u/3` Minimum kinetic energy of system `1/2(3m)(u/3)^(2)=1J` maximum potential energy of system `2J` |
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| 1504. |
The mass of the earth is `9` times that of the mars. The radius of the earth twice that of the mars. The escape velocity of the earth is `12km//s`. The escape velocity on the mars is ........`km//s`A. `4sqrt(2)km//s`B. `2sqrt(2) km//s`C. `6sqrt(2) km//s`D. `8sqrt(2) km//s` |
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Answer» Correct Answer - A `(v_(m))/(v_(e))=sqrt((M_(m))/(M_(e))(R_(e))/(R_(m)))` |
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| 1505. |
Escape velocity on earth is 11.2 km/s . What would be the escape velocity on a planet whose mass is 1000 times and radius is 10 times that of earthA. 112 km/sB. 11.2 km/sC. 1.12 km/sD. 3.7 km/s |
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Answer» Correct Answer - A |
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| 1506. |
A planet has mass `1//10` of that of earth, while radius is `1//3` that of earth. If a person can throw a stone on earth surface height of 90m, then he will be able to throw the stone on that planet to a heightA. `90m`B. `40m`C. `100m`D. `45m` |
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Answer» Correct Answer - C Acceleration due to gravity `g=(GM)/(R^(2))` `:. (g_(plan et))/(g_(earth))=(M_(plan et))/(M_(earth)).((R_("earth"))/(R_("planet")))^(2)=1/10xx(3/1)^(2)=9/10` If a stone is thrown with velocity `u` from the surface of the planet then maximum height `H=(u^(2))/(2g) implies (H_(plan et))/(H_(earth))=(g_(earth))/(g_(plan et))` `implies H_(plan et) =10/9xxH_(earth)=10/9xx90=100 meter` |
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| 1507. |
How many times is escape velocity `(V_(e))` , of orbital velocity `(V_(0))` for a satellite revolving near earthA. `sqrt(2)` timesB. 2 timesC. 3 timesD. 4 times |
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Answer» Correct Answer - C |
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| 1508. |
The least velocity required to throw a body away from the surface of a planet so that it may not return is (radius of the planet is `6.4xx10^(6) m, g=9.8 m//sec^(2)`)A. `9.8xx10^(-3) m//sec`B. `12.8xx10^(3) m//sec`C. `9.8xx10^(3) m//sec`D. `11.2xx10^(3) m//sec` |
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Answer» Correct Answer - D |
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| 1509. |
Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. `GM//12R^(2)`B. `GMm//3R^(2)`C. `GMm//8R`D. `GMm//6R` |
| Answer» Correct Answer - 4 | |
| 1510. |
When a body is close to the surface of the earth, what is the ratio of the escape velocity to the orbital velocity of the body? |
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Answer» The ratio of the escape velocity to the orbital velocity of the body is ve = √e v0. |
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| 1511. |
When a body falls towards earth, earth moves towards the body. Why is earth's motion not noticed? |
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Answer» The motion of the earth is not noticed because the acceleration produced in earth is infinitesimally small, due to larger mass of the earth. |
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| 1512. |
In a uniform circular motion ……………… is constant. A) speed B) force C) gravitation D) weight |
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Answer» Correct option is A) speed |
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| 1513. |
Which of the following quantity always changes in a uniform circular motion ? A) direction of gravitationB) direction of velocity C) mass of the object D) none |
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Answer» B) direction of velocity b) direction of velocity |
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| 1514. |
When a body is in uniform circular motion …………… force acts on it. A) gravitational force B) centrifugal force C) centripetal forceD) normal force |
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Answer» C) centripetal force |
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| 1515. |
Centripetal force actsA) towards the centre of the circle B) away from the centre of the circle C) in a tangential direction D) along the circular path |
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Answer» A) towards the centre of the circle |
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| 1516. |
A satellite is orbiting the earth in a circular orbit of radius `r`. ItsA. kinetic energy veries as `r`B. angular momentum varies as `r^(-1)`C. linear momentum varies as `r^(2)`D. frequency of revolution varies as `r^(-3//2)` |
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Answer» Correct Answer - D Orbital velocity of the satellite is `v=sqrt((GM_(E))/r)` where `M_(E)` is the mass of the earth Kinetic energy, `K=1/2mv^(2)=(GM_(E)m)/(2r)` Where `m` is the mass of the satellite. `K prop 1/r` Hence, option (b) is incorrect. Linear momentum, `p=mv=msqrt((Gm_(E))/r)` `p prop 1/(sqrt(r))` Hence, (c) is incorrect. Frequency of revolution, `v=1/T=1/(2pi)sqrt((GM_(E))/(r^(3)))` `v prop1/(r^(3//2))` Hence, option (d) is correct. |
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| 1517. |
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 0.10 m apart on a horizontal table. What is the gravitational force and potential at the mid-point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable? |
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Answer» Here G = 6.67 x 10-11 Nm2 kg-2 M = 100 kg; R = 0.1 m Distance between two spheres, d = 1.0 m Suppose that the distance of either sphere from the mid-point of the line joining their centre is r. Then, r = d/2 = 0.5 m The gravitational field at the mid-point due to two spheres will be equal and opposite. Hence, the resultant gravitational field at the mid-point = 0 The gravitational potential at the mid-point = (-GM/r) x 2 = (-)6.67 x 10-11 x 100 x 2/0.5 = -2.668 x 10-8 J kg-1 As the net force on the body placed at midpoint zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its, initial position of equilibrium. Hence, the equilibrium attained is unstable. |
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| 1518. |
Mass(m) |
| Answer» The mass of an object is the quantity of matter contained in it. Masss is the measure of the inertia of a body. | |
| 1519. |
Weight (W) |
| Answer» The weight of a body is defined as the force with which the Earth attracts the object. | |
| 1520. |
Observe the figure and anwer the following question: What is the value of universal gravitational constant in SI system? |
| Answer» In SI System, the vallue of `G` is `6.67xx10^(-11) Nm^(2)//kg^(2)` | |
| 1521. |
The gravitational potential energy at the height of h from the ground isA. `(-GMm)/(R+h)`B. `(-GMm_(1))/(R^(2)+h)`C. `(GMm_(1))/(R^(2)+h^(2))`D. `(GMm)/(R^(2)+h)` |
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Answer» Correct Answer - A `(-GMm)/(R+h)` |
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| 1522. |
The orbit of a planet is an ellipse with the Sun at one of the ___________.A. fociB. centreC. middleD. surface |
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Answer» Correct Answer - A foci |
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| 1523. |
Escape velocity |
| Answer» The minimum initial velocity needed by an object projected upwards to overcome Earths gravitational force and not fall back on Earth is called escape velocity. | |
| 1524. |
What would happen if there were no gravity? |
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Answer» There would be no gravitational attraction between any two particles and hence no formation of the solar system, galaxy, etc. |
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| 1525. |
What are the types of forces? |
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Answer» Types of forces are: i. Gravitational force (ii) Electro- magnetic force (iii) Nuclear force (iv) Balanced force (v) Unbalanced force |
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| 1526. |
What would happen if the value of G was twice as large? |
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Answer» The gravitational force between any two particles would become double, also the value of g would become double. |
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| 1527. |
What would happen if the value of G was twice as large? |
| Answer» If the value of G is doubled, then the gravitational force (F) also will get doubled. | |
| 1528. |
Gravitational constant and Gravitational acceleration: |
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Answer» Gravitational constant (G): i. Gravitational constant is the force of attraction between two unit masses placed at unit distance apart form each other. ii. It has SI unit `Nm^(2)//kg^(2)`, while CGS unit is dyne `cm^(2)//g^(2)`. iii. It is always denoted by G. Its. S.I. value is`6.673xx10^(-11)Nm^(2)//kg^(2)`, while in CGS, it is `6.673xx10^(-8)` dyne `cm^(2)//g^(2)`. v.Its value is fixed and does not change with conditions, hence it is called universal constant. Gravitationa acceleration (g) i. The uniform acceleration produced in a freely falling body due to the gravitational force of the Earth is called gravitation acceleration. (ii). It has SI is `m//s^(2)` while CGS unit is `cm//s^(2)` (iii) It is alwyas denoted by g. (iv) It S.I. value is `9.77m//s^(2)`, while is CGS, it is 977 `cm//s^(2)`. (v). The value of g depends on various factors like altitude, depth, shape etc. |
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| 1529. |
In which of the two cases, A or B is the force exerted stronger? |
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Answer» `F=G(m_(1)m_(2))/(d^(2))` Case `A:F_(1)=G(m_(1)m_(2))/(4d^(2))=1/4F` Case `B:F_(2)=G((m_(1)m_(2))/(d^(2)))/9=9F` The force exerted is maximum in case B. |
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| 1530. |
The centre of mass of an object is the point inside or outside the object at which the total mass of the object can be assumed to be concentrated. The centre of mass of spherical object having uniform density is a its geometrical centre. The centre of mass of any object having uniform density is at its centroid. Where can the total mass of an object be assumed to be concentrated? |
| Answer» At the centre of the mass of object. | |
| 1531. |
Weightlessness in space: Space travellers as well as objects in the spacecraft appear to be floating. Why does this happen? Though the spacecradft is at a height from the surface of the earth the value of g there is not zero. In the space station the value of g is only 11% less than its value on the surface of the earth. Thus, the height of a spacecraft is not the reason for their weightlessness. Their weightlessness is caused by their being in the state of free fall. Though the spacecraft is not falling on the earth because of its velocity along the orbit, the only force acting on it is the gravitational force of the earth and therefore it is in a free fall state. As the velocity of free fall does not depend on the properties of an object, the velocity of free falls is the same for the spacecraft, the travellers and the objects in the craft. Thus, if a traveller releases an object from her hand, it will remain stationary with respect to her and will appear to be weightless. Is the value of g zero in the space station? |
| Answer» No, the value of g is only 11% less than its value on the surface of the Earth. | |
| 1532. |
Gravitational waves: Waves are created on the surface of water when we drop a stone into it. Similarly you must have seen the waves generated on a strintg when both its ends are held in hand and it is shaken. Light is also a type of wave called the electromagnetic wave. Gamma rays, X-rays, ultraviolet rays, infrared rays, microwave and radio waves are all different types of electromagnetic waves. Astronomical objects emit these waves and we receive them using our instruments. All our knowledge about the universe has been obtained through these waves.Gravitational waves are a very different type of waves. They have been called the waves on the fabric of space-time. Insteine predicted their existence in 1916. These waves are very weak and it is very difficult to detect them. Scientists have constructed extremely sensitive instruments to detect the gravitational waves emitted by astronomical sources. Among these, LIGO (Laser Interferometrc Gravitational Wave Observatory) is the prominent one. Exactly after hundred years of their prediction, scientists detected these waves coming from an astronomical source. Indian scientist have contributed significantly in this discovery. This discovery has opened a new path to obtain information about the Universe. Is it easy to detect Gravitational waves? |
| Answer» No, these waves are very weak and it is very difficult to detect them. | |
| 1533. |
Gravitational waves: Waves are created on the surface of water when we drop a stone into it. Similarly you must have seen the waves generated on a string when both its ends are held in hand and it is shaken. Light is also a type of wave called the electromagnetic wave. Gamma rays, X-rays, ultraviolet rays, infrared rays, microwave and radio waves are all different types of electromagnetic waves. Astronomical objects emit these waves and we receive them using our instruments. All our knowledge about the universe has been obtained through these waves.Gravitational waves are a very different type of waves. They have been called the waves on the fabric of space-time. Insteine predicted their existence in 1916. These waves are very weak and it is very difficult to detect them. Scientists have constructed extremely sensitive instruments to detect the gravitational waves emitted by astronomical sources. Among these, LIGO (Laser Interferometrc Gravitational Wave Observatory) is the prominent one. Exactly after hundred years of their prediction, scientists detected these waves coming from an astronomical source. Indian scientist have contributed significantly inthis discovery. This discovery has opened a new path to obtain information about the Universe. What is the device, used to detect GRavitational waves? |
| Answer» LIGO(Laser Interferometric Gravitational waves Observatory) is used to detect Gravitational waves. | |
| 1534. |
Gravitational waves: Waves are created on the surface of water when we drop a stone into it. Similarly you must have seen the waves generated on a strintg when both its ends are held in hand and it is shaken. Light is also a type of wave called the electromagnetic wave. Gamma rays, X-rays, ultraviolet rays, infrared rays, microwave and radio waves are all different types of electromagnetic waves. Astronomical objects emit these waves and we receive them using our instruments. All our knowledge about the universe has been obtained through these waves.Gravitational waves are a very different type of waves. They have been called the waves on the fabric of space-time. Insteine predicted their existence in 1916. These waves are very weak and it is very difficult to detect them. Scientists have constructed extremely sensitive instruments to detect the gravitational waves emitted by astronomical sources. Among these, LIGO (Laser Interferometrc Gravitational Wave Observatory) is the prominent one. Exactly after hundred years of their prediction, scientists detected these waves coming from an astronomical source. Indian scientist have contributed significantly in this discovery. This discovery has opened a new path to obtain information about the Universe. What are the waves on the fabric of space time called? |
| Answer» They are called gravitational waves. | |
| 1535. |
Gravitational waves: Waves are created on the surface of water when we drop a stone into it. Similarly you must have seen the waves generated on a strintg when both its ends are held in hand and it is shaken. Light is also a type of wave called the electromagnetic wave. Gamma rays, X-rays, ultraviolet rays, infrared rays, microwave and radio waves are all different types of electromagnetic waves. Astronomical objects emit these waves and we receive them using our instruments. All our knowledge about the universe has been obtained through these waves.Gravitational waves are a very different type of waves. They have been called the waves on the fabric of space-time. Insteine predicted their existence in 1916. These waves are very weak and it is very difficult to detect them. Scientists have constructed extremely sensitive instruments to detect the gravitational waves emitted by astronomical sources. Among these, LIGO (Laser Interferometrc Gravitational Wave Observatory) is the prominent one. Exactly after hundred years of their prediction, scientists detected these waves coming from an astronomical source. Indian scientist have contributed significantly in this discovery. This discovery has opened a new path to obtain information about the Universe. What are the different types of electro magnetic waves? |
| Answer» Gamma rays, X-rays, Ultraviolet rays, infrared rays, microwave and radio waves. | |
| 1536. |
Universal constant of gravitation (G) |
| Answer» Universal constant of gravitation is the force of attraction between two unit masses placed at unit distance apart from each other. | |
| 1537. |
The centre of mass of an object is the point inside or outside the object at which the total mass of the object can be assumed to be concentrated. The centre of mass of spherical object having uniform density is a its geometrical centre. The centre of mass of any object having uniform density is at its centroid. Where is the centre of mass located for a spherical object? |
| Answer» At its geometrical centre. | |
| 1538. |
Observe the figure and answer the following questions: Why gravitational constant is called universal constant? |
| Answer» The value of gravitational constant does not depend upon the nature and size of the bodies. It also does not depend upon the nature of the medium between two bodies, hence it is called universal constant. | |
| 1539. |
Centre of mass |
| Answer» It is the point inside or outside the object at which the total mass of the object can be assumed to be concentrated. | |
| 1540. |
Gravitational acceleration g OR Acceleration due to gravity. |
| Answer» The gravitational force due to earth on a body results in its acceleration. This is called acceleration due to gravity. | |
| 1541. |
Observe the figure and anwer the following question: What will happen to gravitational force if mass of one of the objects is doubled? |
| Answer» If the mass of one of the objects is doubled, then the gravitational force between them also gets doubled. | |
| 1542. |
The force of action and reaction can occur only when two bodies collide against one another.the statememnt isA. trueB. falseC. sometimes true and sometimes falseD. cannot predict. |
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Answer» Correct Answer - B No,the statement is false. |
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| 1543. |
Calculate the distance from the surface of the earth at which the acceleration due to gravity is the same below and above the surface of the earth. |
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Answer» Correct Answer - `(1)/(2)(sqrt(5-1)R` |
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| 1544. |
A body is projected horizontally near the surface of the earth with `sqrt(1.5)` times the orbital velocity .Calculate the maximum heght up which it will rise above the surface of the earth. [Hint : when the velocity of projection exceeds the orbital velocity the path taken is an ellipse At the highest point the radius vector and velocity are right angles Conserve angular momentum and energy.] |
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Answer» Correct Answer - 2R |
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| 1545. |
Choose the wrong option.A. Inertial mass is a measure of difficulty of acceleration a body by an exertnal force whereas the gravitational mass is relevant in determining the gravitational force on it an external mass.B. That the acceleration mass and inertial mass are equal is an experimental result.C. That the acceleration due to gravity on earth is the same for all bodies is due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial massD. Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot. |
| Answer» Correct Answer - D | |
| 1546. |
Which of the following options are correct ?A. Acceleration due to gravity decreases with increasing altitude.B. Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density)C. acceleration due to gravity increases with increasing latitude.D. accleration due to gravity is independent of the mass of the earth |
| Answer» Correct Answer - A::C | |
| 1547. |
particles of masses 2M m and M are resectively at points A , B and C with ` AB = (1)/(2) (BC)` m is much - much smaller than M and at time ` t=0` they are all at rest as given in figure . As subsequent times before any collision takes palce . A. `m` will remainsat restB. `m` will move towards `M`.C. `m` will move towards `2M`D. `m` will have oscillatory motion. |
| Answer» Correct Answer - C | |
| 1548. |
particles of masses 2M m and M are resectively at points A , B and C with ` AB = (1)/(2) (BC)` m is much - much smaller than M and at time ` t=0` they are all at rest as given in figure . As subsequent times before any collision takes palce . 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 Force 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)) le F_(BA)` Hence, `m` will move towards BA (i.e., 2M) |
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| 1549. |
particles of masses 2M m and M are resectively at points A , B and C with ` AB = (1)/(2) (BC)` m is much - much smaller than M and at time ` t=0` they are all at rest as given in figure . As subsequent times before any collision takes palce . 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 Force on B due to `A=F_(BA)=(G(Mm))/((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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| 1550. |
Particles of masses 2M, m and M are respectively at points A, B and C with `AB=(1)/(2)` (BC). M is much smaller than M and at time t = 0, they are all at rest given in the figure. At subsequent times before any collision takes place. A. m will remain at restB. m will move towards MC. m will move towards 2 MD. m will have oscillatory motion |
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Answer» Correct Answer - C Gravitational force on m due to `2M = F_(BA)` `= (G(2M)m)/((AB)^(2))` along `vec(BA)" "` …. (1) Gravitational force on m due to `M = F_(BC)` `=(G(M)m)/((BC)^(2))` along `vec(BC)` `because BC=2AB " " therefore " " F_(BC)=(GMm)/(4AB^(2))" "` ...... (2) Thus from (1) and (2) we find that `F_(BA)gt F_(BC)` `therefore` m will move towards 2M along `vec(BA)`. |
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