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NEET-XII-Physics

11: Gravitation

with Solutions - page 5
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.
    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:
  • #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: