JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    Given that K= energy, V= velocity, T= time. If they are chosen as the fundamental units, then what is dimensional formula for surface tension?   JEE Main Online Paper (Held On 07 May 2012)

    A) \[[K{{V}^{-2}}{{T}^{-2}}]\]           

    B)                        \[[{{K}^{2}}{{V}^{2}}{{T}^{-2}}]\]

    C)                        \[[{{K}^{2}}{{V}^{-2}}{{T}^{-2}}]\]                            

    D)                        \[[K{{V}^{2}}{{T}^{2}}]\]

    Correct Answer: A

    Solution :

                    Surface tension,\[T=\frac{F}{\ell }=\frac{F}{\ell }.\frac{\ell }{\ell }.\frac{{{T}^{2}}}{{{T}^{2}}}\] (As,\[F.\ell =K\](energy)\[\frac{{{T}^{2}}}{{{\ell }^{2}}}={{V}^{-2}}\]) Therefore, surface tension\[=[K{{V}^{-2}}{{T}^{-2}}]\]                


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