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Gravitation Test - 33

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Gravitation Test - 33
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  • Question 1
    1 / -0
    Which Newton's laws describe thrust?
    Solution
    Thrust is a reaction force described quantitatively by Newton's second and third laws. When a system expels or accelerates mass in one direction, the accelerated mass will cause a force of equal magnitude but opposite direction on that system.
  • Question 2
    1 / -0
    Two block of mass $$m_1$$ and $$m_2$$ are kept $$d$$ apart from each other. What happens to the magnitude of the gravitational force $$F$$ acting on the two blocks, if the distance between the center of the masses is halved?
    Solution
    Distance between the masses initially is $$d$$.

    Gravitational force acting between two blocks of mass $$m_1$$ and $$m_2$$ is given by
                   $$F = \dfrac{Gm_1 m_2}{d^2}$$
    Now the distance between them is halved keeping all the parameters same  i.e  $$d' = d/2$$

    Thus gravitational force         $$F' = \dfrac{Gm_1 m_2}{(d/2)^2}  =4F$$

    Thus the force is quadrupled.
  • Question 3
    1 / -0
    Which of the following equations or principle is most suitable to solve given question. 
    The distance between two small sphere is $$d$$ and mass of each sphere is $$ m$$. What force do they apply to each other?
    Solution
    Newton's law of gravitation quantifies the force of attraction between two masses $$m_1,m_2$$ separated by a distance $$d$$ as $$F=\dfrac{Gm_1m_2}{d^2}$$
    Thus the force the spheres apply to each other, according to Newton's Law of Gravitation $$=\dfrac{Gm^2}{d^2}$$
  • Question 4
    1 / -0
    Two block of mass $$m_1$$ and $$m_2$$ are kept apart d from each other. What happen to the magnitude of the force on $$ m_{1}$$ if the mass of $$m_{2}$$ is doubled?
    Solution
    Initial force on $$m_1$$ due to $$m_2$$,               $$F = \dfrac{Gm_1 m_2}{d^2}$$
    Now the mass of $$m_2$$ is doubled keeping all the parameters same  i.e  $$m'_2 = 2m_2$$
    Thus new force on $$m_1$$         $$F' = \dfrac{Gm_1 (2m_2)}{d^2}  =2F$$
    Thus the force also gets doubled.
  • Question 5
    1 / -0
    We have a substance having density x kept in a medium of density y. Then what is its relative density with respect to the medium?
    Solution
    Relative density is defined as the ratio of density of substance to the density of medium.
    $$\therefore$$ Relative density   $$r = \dfrac{x}{y}$$
  • Question 6
    1 / -0
    Relation between density ($$\rho$$), relative density $$r$$, and density of water $$\rho_o$$ is given by:

    Solution
    Relative density is defined as the ratio of density of substance to the density of water.
    $$\therefore$$  Relative density, $$r = \dfrac{\rho}{\rho_o}$$
  • Question 7
    1 / -0
    If we consider the gravitational force F between two objects of masses $$m_1$$ and $$m_2$$  respectively, separated by a distance R , and we double the distance between them, what is the new magnitude of the gravitational force between them? 
    Solution
    According to Sir Newton's gravitational law the magnitude of gravitational force $$F$$ between objects of mass $$m_{1}$$  and $$m_{2}$$  separated by a distance $$R$$ is given by  ,
              $$F=G\dfrac{m_{1}m_{2}}{R^{2}}$$  ............$$eq1$$ ,
    now if the distance between them  is doubled i.e. $$2R$$ ,then the gravitational force will be ,
               $$F'=G\dfrac{m_{1}m_{2}}{\left(2R\right)^{2}}=G\dfrac{m_{1}m_{2}}{4R^{2}}$$ ,
    now by using $$eq1$$ ,
               $$F'=F/4$$
  • Question 8
    1 / -0
    If volume of a steel ball is halved and its mass remains constant, its relative density will be:
    Solution
    Relative density of steel ball  is given by
        $$r = \dfrac{\rho_s}{\rho_w} = \dfrac{m}{V\rho_w}$$            

    where, $$\rho_s$$ is the density of ball and $$\rho_w$$ is the density of water
    $$\implies  r \propto  \dfrac{1}{V}$$          $$(\because m =constant)$$

    Hence, relative density is inversely proportional to the volume of the ball.

    Thus relative density is doubled when the volume of steel ball is halved.
  • Question 9
    1 / -0
    Assuming a person could survive in all of the following locations, where would this person have the most mass?
    Solution
    The weight ($$W=mg$$) of an object depends upon the gravitational acceleration (g). Since the value of $$g$$ is different at different places, the weight of the person will be different at different locations. 
    But, mass is the intrinsic property of the person. It does not change by changing the position. Hence, the mass of the person will be the same in all locations.
  • Question 10
    1 / -0
    A globe of steel has a mass of 12 g and a volume of $$15.2 cm^3$$. Find its relative density wrt water.
    Solution
    Density ($$\rho_s$$) = $$\dfrac{mass(m)}{volume(V)}$$
    Relative density of steel with respect to water=$$\dfrac{\rho_s}{\rho_w}$$
    $$=\dfrac{m/V}{\rho_{w}}$$
    $$=\dfrac{12/15.2}{1}$$
    $$=\dfrac{12}{15.2}$$
    $$=0.79\ g/cm^3$$
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