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Gravitation Test - 79

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Gravitation Test - 79
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  • Question 1
    1 / -0
    A satellite of mass $$'m'$$ is revolving around the earth in an orbit of radius $$10R$$ ($$R$$ is the radius of earth), the minimum energy required to give the satellite so that it can escape the gravitational field of the earth is [mass of earth$$=M_e$$]:
    Solution

  • Question 2
    1 / -0
    At what depth below the surface of the earth, is the value of $$g$$ same as that at a height of $$10\ km$$ from the surface of the earth?
    Solution

  • Question 3
    1 / -0
    If $$K$$ is the kinetic energy of the earth and $$P$$ is the gravitational potential energy of the earth as it revolves around the sun, then which of the following statements is true?
    Solution

  • Question 4
    1 / -0
    If the gravitational acceleration at the Earth's surface is $$9.81\,m{s^{ - 2}}$$, what is its value at a height equal to the diameter of the Earth from its surface?
    Solution

  • Question 5
    1 / -0
    A planet in a distant solar system is $$10$$ times more massive then the earth and its radius is $$10$$ times smaller. Give that escape velocity from the earth is $$11\ km/s$$, the escape velocity from the surface of the plants is
    Solution
    Escape velocity $$= \sqrt{\dfrac{2GM}{R}}$$
    $$\sqrt{\dfrac{2GN}{R}}=11 $$ (given)
    $$M \longrightarrow 10\ M$$ 
    $$R \longrightarrow R/10$$
    Putting value
    $$Vc= \sqrt{\dfrac{2G (10\ m) \times 10}{R}}$$
    $$=10 \times \sqrt{\dfrac{2GM}{R}}$$
    $$=110\ w/s$$
  • Question 6
    1 / -0
    Choose the correct answer:
    The height over the surface of the earth at which the gravitational field of the earth becomes $$\frac{1}{4}$$th of the field at the surface
  • Question 7
    1 / -0
    The kinetic energy needed to project a body of mass m from the earth's surface (radius $$R$$) to infinity is -
    Solution

  • Question 8
    1 / -0
    A rubber ball is dropped from a height of $$5$$cm on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to $$1.8$$m. The ball loses its velocity on bouncing by a factor of?
    Solution
    $$\dfrac { \dfrac { 1 }{ 2 } { { mv }_{ 1 } }^{ 2 } }{ \dfrac { 1 }{ 2 } { { mv }_{ 2 } }^{ 2 } } =\dfrac { mg{ h }_{ 1 } }{ mg{ h }_{ 2 } } =\dfrac { 25 }{ 9 } $$

    $$\dfrac { { { v }_{ 1 } }^{ 2 } }{ { { v }_{ 2 } }^{ 2 } } =\dfrac { 25 }{ 9 } $$

    $$\dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { 5 }{ 3 } $$
  • Question 9
    1 / -0
    A man slides down a light rope whose breaking strength $$\eta$$ times the weight $$(\eta < 1)$$. The maximum acceleration of the man so that the rope just breaks is:-
    Solution
    Breaking strength $$=$$ $$-$$ maximum tension 
    Given that
    $$T=\eta \times $$ weight of man $$=\eta \times mg$$
    From Newton's law of motion
    $$mg-T=ma$$
    $$mg-\eta mg=ma$$
    $$mg\left( 1-\eta  \right) =ma$$
    $$a=\left( 1-\eta  \right) g$$
    Given that, $$\eta <1$$. So, acceleration is downwards because $$\left( 1-\eta  \right) >0$$
    So,  $$a=\left( 1-\eta  \right) g$$ option (A) correct.
  • Question 10
    1 / -0
    The depth at which the effective value of acceleration due to gravity is one fourth that of its value on the earth's surface is 
    Solution

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