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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 26

  • question_answer
    A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity of \[2\sqrt{2}\,rad.{{s}^{-1}}\]. The radius of cylinder must be \[\left( Take g = 10 m{{s}^{-}}^{2} \right)\]

    A) 5 cm                

    B) 0.5 cm

    C) \[\sqrt{10}\] cm 

    D)        5 m

    Correct Answer: D

    Solution :

    \[\nu =\sqrt{\frac{2gh}{1+\frac{I}{m{{r}^{2}}}}}\,\,=\,\,\sqrt{\frac{2\times 10\times 3}{1+\frac{m{{r}^{2}}}{2\times m{{r}^{2}}}}}\,\,=\,\,\sqrt{\frac{2\times 10\times 3}{\frac{3}{2}}}\,\,=\,\,\sqrt{40}\]


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