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Gravitation Test - 53

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Gravitation Test - 53
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  • Question 1
    1 / -0
    Two particles of mass $$m_1$$ and $$M_2$$, approach each other due to their mutual gravitational attraction only. Then
    Solution
    The force acting on each particle due to gravitation=$$F=\dfrac{Gm_1m_2}{r^2}$$
    Acceleration of mass $$m_1=\dfrac{F}{m_1}=\dfrac{Gm_2}{r^2}$$

    Acceleration of mass $$m_2=\dfrac{F}{m_2}=\dfrac{Gm_1}{r^2}$$
    Hence correct answer is option C.
  • Question 2
    1 / -0
    Dimensional formula of universal gravitational constant $$G$$ is-
    Solution

  • Question 3
    1 / -0
    Suppose that you have a mass of $$70\ kg$$. How much mass must another object have in order for your body and the other object to attract each other with a force of $$1
    \ N$$ when separated by $$10\ m$$?
    Solution
    Given:
    Mass of object-1 is $$m_1=70\ kg$$
    Distance between the otwo objects $$r=10\ m$$
     Gravitational force $$F=1\ N$$
    Mass of object-2 is $$m_2=?$$
    We know from Newton's law of gravitation that $$F=\dfrac{Gm_1m_2}{r^2}$$
    $$1=\dfrac{6.67\times 10^{-11}\times 70\times m_2}{10^2}$$
    $$m_2=2.14\times 10^{10} kg$$
  • Question 4
    1 / -0
    If earth's radius were to hypothetically shrink by $$1\%$$, the value of $$G$$ would:
    Solution
    According to the universal law of gravitation, every object in the universe attracts every other objects with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects.

    $$F=G\dfrac{m_1m_2}{d^2}$$

    where $$G$$ is the constant of proportionality and is called the universal gravitation constant. It does not depend on the radius.
    So, if the earth's radius shrinks by $$1\%$$, the value of $$G$$ would remain the same.
  • Question 5
    1 / -0
    The relationship between acceleration due to gravity($$g$$) and universal gravitational constant($$G$$) may be represented as:
    ($$M$$ and $$R$$ are the mass and radius of the earth respectively )

    Solution
    Gravitational force acting on the mass $$m$$  $$F_g = \dfrac{GMm}{R^2}$$

    But $$F_g = mg$$

    $$\therefore$$ $$mg = \dfrac{GMm}{R^2}$$

    We get $$g = \dfrac{GM}{R^2}$$
  • Question 6
    1 / -0
    How the gravitational constant will change if a brass plate is introduced between two bodies?
    Solution
    Universal Gravitational Constant (G) remains constant irrespective of the medium between two bodies. It has a constant value of $$6.67  \times  10^{-11} Nm^{2}kg^{-2}$$
  • Question 7
    1 / -0
    The moon's radius is $$\dfrac{1}{4}$$ that of the earth and its mass is $$\dfrac{1}{80}$$ times that of the earth. If $$g$$ represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is 
    Solution
    The gravitational acceleration on surface of a planet is given as:
    $$g=\dfrac{GM}{R^2}$$

    Hence $$\dfrac{g_m}{g_e}=\dfrac{M_m}{M_e}.\dfrac{R_e^2}{R_m^2}$$ $$=\dfrac{1}{80}.\dfrac{4^2}{1}$$ $$=\dfrac{1}{5}$$

    $$\implies g_m=\dfrac{g}{5}$$
  • Question 8
    1 / -0
    An astronaut on a strange planet finds that acceleration due to gravity is twice as that on the surface of Earth. Which of the following could explain this?
    Solution
    The expression for acceleration due to gravity on the surface of a planet is given by $$g=\dfrac{GM}{R^2}.$$
    if g on the other planet is double then this will be satisfied by the only condition in option C.
    in option A, g on other planet is half that of the earth.
    in option B,  g on other planet is half that of the earth.
    in option D, g on other planet is four times that of the earth.
    so the best possible answer is option C.
  • Question 9
    1 / -0
    Which of the following hypotheses was made by Newton?
    Solution
    The law of gravitation states that the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between
    them.
    $$F_g = \dfrac{Gm_1 m_2}{r^2}$$
    where $$G$$ is the gravitational constant.
  • Question 10
    1 / -0
    A bodyweight $$5 N$$ in the air and $$2 N$$ when immersed in a liquid. The buoyant force is:
    Solution
    Given, The weight of the body in the air is $$=5N,$$
    When immersed in water, the weight of the body $$=2N$$
    In water, there will be a buoyancy force Which opposes the weight of the body.
    Therefore the buoyant force is equal to $$=5N-2N=3N$$
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