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Gravitation Test - 63

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Gravitation Test - 63
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following statements should be correct for a rocket fired from the earth towards the moon ?
    Solution

  • Question 2
    1 / -0
    A body of mass $$5\ kg$$ is cut into two parts of masses (a) $$\dfrac {m}{4}; \dfrac {3m}{4}$$ (b) $$\dfrac {m}{7}; \dfrac {5m}{7}$$ (c) $$\dfrac {m}{2}; \dfrac {m}{2}$$ (d) $$\dfrac {m}{5}; \dfrac {4m}{5}$$. When these two pieces are kept apart by certain distance; In which case the gravitational force acting is maximum?
    Solution

  • Question 3
    1 / -0
    The weight of a person reduces to one-sixth on the Moon than on the Earth What changes will you notice in the mass of a person on the Moon in comparison to the Earth?
    Solution
    The mass of an object is the measure of its inertia. It remains the same whether the object is on the earth, the moon, or even in outer space. Thus, the mass of an object is constant and does not change from place to place.
    The weight of an object is the force with which it is attracted towards the earth.
    $$Weight=m \times g$$
    where $$m=$$ mass and $$g=$$ acceleration due to gravity.
  • Question 4
    1 / -0
    A balloon filled with air is weighed(W) so that it just floats in water as shown in the figure. When it is further pushed by a short distance in water, it will.

    Solution
    Assuming that "barely able to float" means that it still IS able to float, then it's density is still less than water's and it will return to the surface after being pushed below the surface. 
    However, if by "barely able to float" the questions means that the weighted balloon's density is the same as water's, then it will not return to the surface of the water; it will just continue to float at the depth.
    Hence Option A is correct.
  • Question 5
    1 / -0
    The moon's radius is $$1/4$$ times that of the earth and its mass $$1/80$$ times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, then on the surface of the moon its value is :
    Solution
    Acceleration due to gravity on the earth's surface is $$g=\dfrac{GM_e}{R_e^2}$$
    Acceleration due to gravity on the moon's surface is $$g_m=\dfrac{GM_m}{R_m^2}$$
    Given, $$R_m=\dfrac{R_e}{4}$$ or $$\dfrac{R_m}{R_e}=\dfrac{1}{4}$$ and $$M_m=\dfrac{M_e}{80}$$ or $$\dfrac{M_m}{M_e}=\dfrac{1}{80}$$
    So, $$\dfrac{g_m}{g}=\dfrac{M_m}{M_e}\times (\dfrac{R_e}{R_m})^2=(1/80)\times (4)^2=1/5$$
    or $$g_m=g/5$$
  • Question 6
    1 / -0
    If $$R$$ is the radius of a planet, $$g$$ is the acceleration due to gravity and $$ G $$ is the gravitational constant then find the density of the planet.
    Solution
    Acceleration due to gravity is given by,  $$g=\cfrac { GM }{ { R }^{ 2 } } $$..........$$(1)$$
    where G is the gravitational constant, M is the mass of the planet, R is the radius of the planet.
    Now $$\text{Mass} = \text{Volume }\times\text{density of planet} $$
     M $$=\cfrac { 4 }{ 3 } \pi { R }^{ 3 }\times \rho$$ ,  where $$\rho$$  is the Density of the planet
    Equation $$(1)$$ becomes
    $$g=\cfrac { G\cfrac { 4 }{ 3 } \pi { R }^{ 3 }\rho }{ { R }^{ 2 } } \\ \rho=\cfrac { 3 g}{ 4G\pi R } $$
  • Question 7
    1 / -0
    A mass is suspended from a spring having constant 'K' is displaced vertically and released, it oscillates with period 'T'. The weight of the mass suspended is ($$g =$$ gravitational acceleration)
    Solution

  • Question 8
    1 / -0
    Suppose universal gravitational constant of starts decrease, then:
    Solution

  • Question 9
    1 / -0
    The radius of the Earth is about $$6400 km$$ and that of Mars is about $$3200 km$$. The mass of the Earth is about $$20$$ times the mass of Mars. An object weighs $$500 N$$ on the surface of Earth. Its weight on the surface of Mars would be:
    Solution

  • Question 10
    1 / -0
    Two masses $$m$$ and $$M$$ are kept at a distance $$r$$ from each other. The ratio of the force exerted on $$m$$ due to $$M$$ and that of $$M$$ due to $$m$$ is equal to: 
    Solution

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