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Gravitation Test - 17

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Gravitation Test - 17
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  • Question 1
    1 / -0
    The gravitational field is a conservative field. The work done in this field by moving an object from one point to another
    Solution
    A conservative  force is a type of force wherein there is no net work done during its motion in any closed loop.
    When we throw a ball up there is negative work done as it moves against the force of gravity and during the fall the gravity is in the positive direction. The resultant work done is zero and the force of gravity is path independent.
    Thus, gravitational field is a conservative field and  work done depends on the end points only.
  • Question 2
    1 / -0
    Which of the following quantities remain constant in a planetary motion, when seen from the surface of the sun?
    Solution
    When viewed from the inertial frame of the sun, the planets seem to execute elliptical orbital motion with changing ($$K.E$$) (Kepler's first law) but constant angular momentum (due to the force being central).
  • Question 3
    1 / -0
    The escape velocity from the earth for a rocket is 11.2 km/sec. Ignoring the air resistance, the escape velocity of 10 mg grain of sand from the earth will be
    Solution
    Escape velocity on a planet depends on the mass and radius of the planet alone. Hence, both for the rocket and the sand grain, escape velocity is 11.2 km/sec.
  • Question 4
    1 / -0
    A gravitational field is present in a region. A point mass is shifted from $$A$$ to $$B$$, along different paths shown in the figure. If $$W_{1}$$ , $$W_{2}$$ and $$W_{3}$$ represent the work done by gravitational force for respective paths, then

    Solution
    Work done is independent of the distance traveled or the path chosen. It depends only on the displacement. 
    Since for all the paths given, the displacement is same, the work done by gravitational field in each case is equal.
  • Question 5
    1 / -0
    The escape velocity of a body depends upon its mass as
    Solution
    Escape speed of a body from Earth's surface is given by: $$v_{min}= \sqrt {2gR}$$
    As we can see from the above equation, there is no 'mass' term. Implies it can be written as $$m^{0}$$.
    So, the escape speed of a body is independent of its mass.
  • Question 6
    1 / -0
    The earth retains its atmosphere. This is due to
    Solution
    The earth retains its atmosphere. This is due to the escape velocity being greater than the mean speed of the molecules of the atmospheric gases.
  • Question 7
    1 / -0
    If $$g$$ on the surface of the Earth is $$9.8$$ $$ms^{-2}$$, then it's value at a depth of $$3200$$ $$km$$ (Radius of the earth $$ =  6400$$ $$km$$) is
    Solution
    The value of gravity changes as we move away from or towards the centre of the Earth.
    This is given by: $$g_{R} =\dfrac{GM}{ R^{2} }$$, where $$M$$ is the mass of a planet of radius $$R$$
    So, $$M=\dfrac{4}{3}\pi R^{3}\rho$$ ; substituting in above equation
     $$\Rightarrow g_R = G \dfrac{4}{3} \pi \rho \times R $$
    Since we want the value of $$g$$ at depth($$d$$) from the Earth's surface, we replace $$R$$ by $$(R-d)$$
    $$\Rightarrow g_d = G \dfrac{4}{3} \pi \rho \times (R-d) $$
    $$\Rightarrow \dfrac{g_{R}}{ g_{d} } = \dfrac{R}{R-d} $$
    $$\Rightarrow  \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{d}{R})$$

    The given depth in the problem is $$d = 3200$$  km, substituting we get,
    $$ \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{3200}{6400})$$
    $$ {g_{d}}=9.8\times(1- \dfrac{3200}{6400})$$
    $$ {g_{d}}=9.8/2$$
    $$ {g_{d}}=4.9 m s^{-2} $$
  • Question 8
    1 / -0
    The value of quantity 'G' in the law of gravitation:
    Solution
    The value of G is a constant like other constants and is independent of factors such as medium, temperature pressure etc. The  value of gravitational constant of the earth is independent on  the mass and radius of the earth.
  • Question 9
    1 / -0
    The escape velocity from the earth for a rocket is $$11.2$$ km/s ignoring air resistance. The escape velocity of $$10$$ mg grain of sand from the earth will be
  • Question 10
    1 / -0
    The acceleration due to gravity at a depth of $$1600km$$ inside the earth is
    Solution
    The value of $$g$$ at a depth $$d$$ is given by:
    $$g^{'} = g \left ( 1-\dfrac{d}{R}\right)=9.81 \left(1-\dfrac {1600}{6400} \right) =9.81\times \left( \dfrac{3}{4} \right)=7.35$$ $$ ms$$$$^{-2}$$
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