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NEET-XII-Physics

11: Gravitation

with Solutions - page 6
Qstn# iv-20 Prvs-QstnNext-Qstn
  • #20
    The gravitational potential in a region is given by V = 20 N kg-1 (x + y). (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x, y). Leave your answer in terms of the unit vectors
    i→, j→, k→. (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin. (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x, y). Leave your answer in terms of the unit vectors
    i→, j→, k→. (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin.
    Ans : (a) `` V=\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\left(x+y\right)``
    `` \left[\frac{\,\mathrm{\,GM\,}}{\,\mathrm{\,R\,}}\right]=\left[\frac{{\,\mathrm{\,MLT\,}}^{-2}}{\,\mathrm{\,M\,}}\right]\left[\,\mathrm{\,L\,}\right]``
    `` \Rightarrow \left[\frac{{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-2}{\,\mathrm{\,M\,}}^{1}}{\,\mathrm{\,L\,}}\right]=\left[\frac{{\,\mathrm{\,ML\,}}^{2}{\,\mathrm{\,T\,}}^{-2}}{\,\mathrm{\,M\,}}\right]``
    `` \Rightarrow \left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]=\left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]``
    `` \therefore \,\mathrm{\,LHS\,}=\,\mathrm{\,RHS\,}`` (b) The gravitational field at the point (x, y) is given by `` {\stackrel{\to }{\,\mathrm{\,E\,}}}_{\left(x,y\right)}=-20\left(\frac{\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}``. (c) `` \stackrel{\to }{\,\mathrm{\,F\,}}=\stackrel{\to }{\,\mathrm{\,E\,}}m``
    `` =0.5\,\mathrm{\,kg\,}\left[-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}\right]``
    `` =-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{i}-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{j}``
    `` \therefore \left|\stackrel{\to }{\,\mathrm{\,F\,}}\right|=\sqrt{\left(100\right)+\left(100\right)}``
    `` =10\sqrt{2}\,\mathrm{\,N\,}``
    Page No 227: (a) `` V=\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\left(x+y\right)``
    `` \left[\frac{\,\mathrm{\,GM\,}}{\,\mathrm{\,R\,}}\right]=\left[\frac{{\,\mathrm{\,MLT\,}}^{-2}}{\,\mathrm{\,M\,}}\right]\left[\,\mathrm{\,L\,}\right]``
    `` \Rightarrow \left[\frac{{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-2}{\,\mathrm{\,M\,}}^{1}}{\,\mathrm{\,L\,}}\right]=\left[\frac{{\,\mathrm{\,ML\,}}^{2}{\,\mathrm{\,T\,}}^{-2}}{\,\mathrm{\,M\,}}\right]``
    `` \Rightarrow \left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]=\left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]``
    `` \therefore \,\mathrm{\,LHS\,}=\,\mathrm{\,RHS\,}`` (b) The gravitational field at the point (x, y) is given by `` {\stackrel{\to }{\,\mathrm{\,E\,}}}_{\left(x,y\right)}=-20\left(\frac{\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}``. (c) `` \stackrel{\to }{\,\mathrm{\,F\,}}=\stackrel{\to }{\,\mathrm{\,E\,}}m``
    `` =0.5\,\mathrm{\,kg\,}\left[-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}\right]``
    `` =-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{i}-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{j}``
    `` \therefore \left|\stackrel{\to }{\,\mathrm{\,F\,}}\right|=\sqrt{\left(100\right)+\left(100\right)}``
    `` =10\sqrt{2}\,\mathrm{\,N\,}``
    Page No 227:
  • #20-a
    Show that the equation is dimensionally correct.
    Ans : `` V=\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\left(x+y\right)``
    `` \left[\frac{\,\mathrm{\,GM\,}}{\,\mathrm{\,R\,}}\right]=\left[\frac{{\,\mathrm{\,MLT\,}}^{-2}}{\,\mathrm{\,M\,}}\right]\left[\,\mathrm{\,L\,}\right]``
    `` \Rightarrow \left[\frac{{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-2}{\,\mathrm{\,M\,}}^{1}}{\,\mathrm{\,L\,}}\right]=\left[\frac{{\,\mathrm{\,ML\,}}^{2}{\,\mathrm{\,T\,}}^{-2}}{\,\mathrm{\,M\,}}\right]``
    `` \Rightarrow \left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]=\left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]``
    `` \therefore \,\mathrm{\,LHS\,}=\,\mathrm{\,RHS\,}``
  • #20-b
    Find the gravitational field at the point (x, y). Leave your answer in terms of the unit vectors
    i→, j→, k→.
    Ans : The gravitational field at the point (x, y) is given by `` {\stackrel{\to }{\,\mathrm{\,E\,}}}_{\left(x,y\right)}=-20\left(\frac{\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}``.
  • #20-c
    Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin.
    Ans : `` \stackrel{\to }{\,\mathrm{\,F\,}}=\stackrel{\to }{\,\mathrm{\,E\,}}m``
    `` =0.5\,\mathrm{\,kg\,}\left[-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}\right]``
    `` =-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{i}-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{j}``
    `` \therefore \left|\stackrel{\to }{\,\mathrm{\,F\,}}\right|=\sqrt{\left(100\right)+\left(100\right)}``
    `` =10\sqrt{2}\,\mathrm{\,N\,}``
    Page No 227: