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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    A uniform rod of length \[l\], hinged at the lower end is free to rotate in a vertical plane. Initially, the rod is held vertically and then released. The angular acceleration of the rod when it makes an angle of\[45{}^\circ \]with the vertical is

    A)  \[\frac{2g}{3\sqrt{2}l}\]                               

    B)  \[\frac{3g}{4\sqrt{2}l}\]

    C)  \[\frac{\sqrt{3}g}{2l}\]                 

    D)  \[\frac{3g}{2\sqrt{2\,}l}\]

    Correct Answer: D

    Solution :

     \[Mg\cdot \frac{l}{2}\sin \theta =I\alpha \] \[I=\frac{M{{l}^{2}}}{3}\] \[\Rightarrow \]               \[\alpha =\frac{mgl\,\sin \theta }{2}\times \frac{3}{m{{l}^{2}}}\] \[=\frac{3g}{2{{l}^{2}}}\sin \theta \] At \[\theta =45{}^\circ ,\]\[\alpha =\frac{3g}{2\sqrt{2}l}\]


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