JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    If speed (V), acceleration and force (F) are considered as fundamental units, the dimension of Young's modulus will be [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[{{V}^{-2}}{{A}^{2}}{{F}^{-2}}\]                        

    B) \[{{V}^{-2}}{{A}^{2}}{{F}^{2}}\]

    C) \[{{V}^{-4}}{{A}^{2}}F\]                 

    D) \[{{V}^{-4}}{{A}^{-2}}F\]

    Correct Answer: C

    Solution :

    Young's modulus,\[Y=\frac{F/A}{\Delta l/l}\] Dimensionally,\[Y=\frac{F}{{{(length)}^{2}}}=\frac{F}{{{(speed)}^{2}}{{(time)}^{2}}}\] \[=\frac{F}{{{(speed)}^{2}}{{\left( \frac{Speed}{Acceleration} \right)}^{2}}}\] \[[Y]=[F\,{{V}^{-4}}{{A}^{2}}]\]


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