JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be :- [JEE Main 8-4-2019 Afternoon]

    A) \[{{S}^{3/2}}{{I}^{1/2}}{{h}^{0}}\]              

    B) \[{{S}^{1/2}}{{I}^{1/2}}{{h}^{0}}\]

    C) \[{{S}^{1/2}}{{I}^{1/2}}{{h}^{-1}}\] 

    D)   \[{{S}^{1/2}}{{I}^{3/2}}{{h}^{-1}}\]

    Correct Answer: B

    Solution :

    \[p=k\,{{s}^{s}}{{I}^{b}}{{h}^{c}}\] where k is dimensionless constant \[ML{{T}^{-1}}={{(M{{T}^{-2}})}^{a}}{{(M{{L}^{2}})}^{b}}{{(M{{L}^{2}}{{T}^{-1}})}^{c}}\] \[a+b+c=1\] \[2b+2c=1\] \[2ac=1\] \[a=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,b=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,c=0\] \[{{s}^{1/2}}{{I}^{1/2}}{{h}^{0}}\]


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