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Gravitation Test - 8

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Gravitation Test - 8
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  • Question 1
    1 / -0
    If the earth is replaced by a small point sized particle of the same mass, it is the geodesic centre. How will the force of gravity change at the present surface of the earth? (Assuming the earth to be a perfect sphere)
    Solution
    While calculating the gravity at any point, the distance of point is considered from the centre of the earth. Hence, even if the earth is replaced by a small point sized particle of the same mass, the distance of a point at the present surface of the earth remains the same. Therefore, the gravity will remain the same.
  • Question 2
    1 / -0
    What causes tides?
    (i) Attraction of sun
    (ii) Attraction of moon
    (iii) Attraction of earth
    Solution
    Gravitation attraction of the sun and the moon on the oceans of the earth are responsible for tides.
    Thus, option (3) is correct.
  • Question 3
    1 / -0
    If somehow, the distance between the Sun and Earth is doubled, then the gravitational force between them will become
    Solution
    F = Gm1m2/r2, where F = gravitation force between the Sun and Earth, G = universal gravitation constant, m1 = mass of the Sun, m2 = mass of Earth and r = distance between the Sun and Earth.
    Now, when the distance between the two bodies is doubled, the distance becomes 2r and the force is denoted by F'.
    F' = Gm1m2/(2r)2
    F' = F/4
    Thus, option 4 is correct.
  • Question 4
    1 / -0
    At what height above the Earth's surface is the acceleration due to gravity 1% less than its value at the surface?
    Solution
  • Question 5
    1 / -0
    A planet has twice the values of mass and radius as of Earth. Acceleration due to gravity on the surface of this planet is
    Solution
    g = GM/R2
    g' = G(2M)/(2R)2
    g' = g/2 = (9.8/2) m/s2 = 4.9 m/s2
  • Question 6
    1 / -0
    Two planets have radii r1 and r2 and densities d1 and d2. The ratio of acceleration due to gravity on them will be
    Solution
    g = GM/
    g1 = G x d1 x (4/3 )/
    g1 = Gd1 x (4/3)
    Similarly,
    g2 = Gd2 x (4/3)
    g1/g2 = r1d1/r2d2
  • Question 7
    1 / -0
    The rate of fall of g on going up the earth's surface is ______ the rate when going down the earth's surface.
    Solution
    Moreover at the center of the earth, the force of gravity is maximum. So, the rate of fall of gravity going up the earth's surface is double the rate when going down the earth's surface.
  • Question 8
    1 / -0
    The weights of an object in a coal mine, at sea level and on the top of a mountain are W1, W2 and W3, respectively. Which of the following relations holds true?
    Solution
    W1 < W2 > W3 is correct.
    The weight of an object in a coal mine is lesser than the weight of that object at sea level. The weight of an object is lesser on the top of a mountain than at sea level.
  • Question 9
    1 / -0
    An object shoots off perpendicularly from the earth with a velocity 4/5 times the escape velocity. What will be the maximum height from the surface of the earth up to which the object will rise? (Given: R = radius of earth)
    Solution
  • Question 10
    1 / -0
    Which of the following statements are true?
    (i) Orbital velocity is more than escape velocity.
    (ii) Escape velocity is times the orbital velocity.
    (iii) Orbital velocity increases with radius.
    (iv) A body with a velocity equal to the orbital velocity will orbit the planet in a circular path.
    Solution
    (i) Escape velocity is times the orbital velocity.
    (ii) A body with a velocity equal to the orbital velocity will orbit the planet in a circular path.
  • Question 11
    1 / -0
    Escape velocity of a projectile from the surface of earth is about
    Solution
    Escape velocity of a projectile from the surface of earth is about 11.2 km/s
  • Question 12
    1 / -0
    A pendulum has a velocity v at the lowest point. What is the maximum height reached by it?
    Solution
    Let d be the maximum height reaches by the pendulum
    v2 = 2gd
    d = v2/2g
  • Question 13
    1 / -0
    An object is orbiting at double its original radius. Its orbital velocity has changed by a factor of
    Solution
    v =
    v' =
  • Question 14
    1 / -0
    The mass of a satellite increases by a factor of two. The orbital velocity
    Solution
    The mass of a satellite increases by a factor of two. The orbital velocity remains unchanged because orbital velocity of satellite does not depend upon the mass of the satellite.
  • Question 15
    1 / -0
    Inside a satellite, orbiting very close the earth's surface, water does not fall out of a glass when it is inverted. Which of the following is the best explanation for this?
    Solution
    The water and the glass have the same acceleration, equal to g, towards the centre of the earth. Hence, there is no relative motion between them.
  • Question 16
    1 / -0
    Which of the following is not a statement of Kepler's Law?
    Solution
    Planets move around the sun in a circular orbit is not a statement of Kepler's Law.
  • Question 17
    1 / -0
    A planet goes around the sun twice as many times as earth in a given time interval.
    Solution
    As Kepler's third law states that square of the time period is directly proportional to the cube of the semi major axis .



    hence lesser is the time period of revolution lesser is the radius of the orbit.
  • Question 18
    1 / -0
    A satellite going round the earth in a circular orbit loses some energy due to a collision. If its speed is v and distance from the earth is d, which of the following is correct?
    Solution
  • Question 19
    1 / -0
    When a planet moves around the sun, its
    Solution
    When a planet moves around the sun, its areal velocity is constant
  • Question 20
    1 / -0
    When a body moves with high speed, albeit, much below the speed of light, its mass
    Solution
    In Newtonian physics, the masss of a body is not dependent on the speed of the body.
    Thus, option (1) is correct.
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