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Gravitation Test - 4

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Gravitation Test - 4
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  • Question 1
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    The direction of the universal gravitational force between particles of masses m1 and m2 is:

    Solution

    since gravitational force is attractive in nature, so if there are two particles of m1 & m2 .

    So One Force will be on m1 which will be directed towards m 2. i.e 

    And other force wiil be on m2 which will be directed towards m 1 i.e

    Clearly, It can be seen that  Because thsese forces are attracted to each other and direction of each force is opposite to other one.

     

  • Question 2
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    The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is

    Solution

    According to Shell's theorem, If a particle of mass m is located outside a spherical shell of mass M at, for instance, point P, the shell attracts the particle as though the mass of the shell were concentrated at its centre. Thus, as far as the gravitational force acting on a particle outside the shell is concerned, a spherical shell acts no differently from the solid spherical distributions of mass.

  • Question 3
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    If the radius of Earth decreases by 1% and its mass remains the same, then the acceleration due to gravity
  • Question 4
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    The gravitational force on point mass m1 due to point mass m2, m3 & m4 is:

    Solution

    According to properties of gravitational force: force between the particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to a number of particles is the resultant of forces due to individual particles i.e. 

     

  • Question 5
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    Two bodies of masses m and M are placed at a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is 
  • Question 6
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    A ‘central’ force is always directed

    Solution

    The motion of a particle under a central force F always remains in the plane defined by its initial position and velocity.This may be seen by symmetry. Since the position r, velocity v and force F all lie in the same plane, there is never an acceleration perpendicular to that plane, because that would break the symmetry between "above" the plane and "below" the plane.

    To demonstrate this mathematically, it suffices to show that the angular momentum of the particle is constant. This angular momentum L is defined by the equation

    L = r × p = r × m v

     

  • Question 7
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    The square of the angular velocity of a planet around the Sun is proportional to (where R is the radius of the orbit of the planet around the Sun)
  • Question 8
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    If the distance between the Sun and the Earth is r, then the angular momentum of Earth around the Sun is proportional to
  • Question 9
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    According to Kepler’s Law of periods, The  _______________ of the time period of revolution of a planet is proportional to the cube of the  ___________ of the ellipse traced out by the planet

    Solution

    Kepler's 3rd Law: T= a3. Kepler's 3rd law is a mathematical formula. It means that if you know the period of a planet's orbit (T = how long it takes the planet to go around the Sun), then you can determine that planet's distance from the Sun (a is the length of the semimajor axis of the planet's orbit)

  • Question 10
    1 / -0
    Directions: The following question has four choices out of which ONLY ONE is correct.

    The gravitational field in a region is given by I = (3i + 4j) N / Kg. Work done by this field is zero when a particle is moving along the line. Find the line.
  • Question 11
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    According to Kepler’s Law of orbits:

    Solution

    The orbit of a planet around the Sun (or of a satellite around a planet) is not a perfect circle. It is an ellipse—a “flattened” circle. The Sun (or the centre of the planet) occupies one focus of the ellipse. A focus is one of the two internal points that help determine the shape of an ellipse. The distance from one focus to any point on the ellipse and then back to the second focus is always the same.

  • Question 12
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    According to Kepler’s Law of areas,

    Solution

    Kepler law of areas describes the speed of a planet travelling in an elliptical orbit around the sun. A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit. So, statement of Kepler's law of area is "The line joining the sun to the planet sweeps out equal areas in equal interval of time". i.e. areal velocity is constant.

  • Question 13
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    The angular velocity of a planet revolving in an elliptical orbit around the sun __________ when it comes near the sun
  • Question 14
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    Two satellites of Earth, S1 and S2, are moving in the same orbit. The mass of S1 is four times the mass of S2. Which of the following statements is true?
  • Question 15
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    A tunnel is dug along a diameter of the earth. What is the quantity of the force acting on a particle of mass m placed in the tunnel at a distance r from the surface of earth? Assume Mass of the earth is M and the radius of earth is R
  • Question 16
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    Four particles of a mass M are located at the vertices of a square with side L. What is the gravitational potential due to this at the centre of the square?
  • Question 17
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    A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure.






    The sphere with cavity now applies a gravitational force F2 on the same particle. The ratio of F2 : F1 is
  • Question 18
    1 / -0
    Directions: The following question has four choices out of which ONLY ONE is correct.

    Three particles each of mass 'm' are placed at the corners of an equilateral triangle of side 'b'. What will be the gravitational potential energy of the system of particles?
  • Question 19
    1 / -0

    Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude

    Solution

    According to Newton's law of universal gravitation, the force of attraction between two objects is directly proportional to the product of two masses and inversely proportional to the square of the distance between them

  • Question 20
    1 / -0

    To find the resultant gravitational force acting on the particle m due to a number of masses we need to use:

    Solution

    According to properties of gravitational force, gravitational force between the particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles is the resultant of forces due to individual particles i.e. 

     

  • Question 21
    1 / -0
    Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational attraction between them is proportional to
  • Question 22
    1 / -0
    Directions: The following question has four choices out of which ONLY ONE is correct.

    The gravitational potential energy of a body at a distance 'r' from the centre of earth is U. What will be its weight at a distance R from the centre of the Earth?
  • Question 23
    1 / -0
    Directions: The following question has four choices out of which ONLY ONE is correct.

    Four particles of mass m each are placed at four corners of a square of side a. What is the work needed to be done on this system to increase the side of the square to 2a?
  • Question 24
    1 / -0
    Directions: The following question has four choices out of which ONLY ONE is correct.

    If a body is raised from the surface of Earth upto height 'R', what is the change in potential energy?
  • Question 25
    1 / -0

    The area ΔA swept out by a planet of mass m in time interval Δt is related to the angular momentum by:

    Solution

    The law of areas can be understood as the consequence of conservation of angular momentum which is valid for any central for A central force is such that the force on the planet is along the vector joining the sun and the planet. Let the sun be at the origin and let the position and momentum of the planet be denoted by r and p  respectively. Then the area swept out by the planet of mass m in the time interval Δt is given by ΔA = ½  (r × vΔt).

    Hence ΔA/Δt    =½ (r × p)/m, (since  v=p/m)

    =    L / (2 m) 

    where v is the velocity,  where v is the velocity,  L is the angular

    momentum equal to   ( r  ×  p  ).  For a central force, which is directed along r, L is a constant

     

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