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JEE Main & Advanced Sample Paper JEE Main Sample Paper-22

  • question_answer
     A bead of mass m is attached to a spring of natural length \[\left( \frac{d+x}{d-x} \right)W\] with other end of it fixed at O. The spring moves in a track of which part ABC is semicircular of radius R and path CDA varies from \[R\,to\frac{R}{2}\] and then again to R. The bead is in equilibrium at D and starts moving downward. Find the ratio of normal reaction on the bead to the centrifugal force in the bottom most position.

    A)   2                               

    B)  \[\frac{1}{2}\]

    C)  1.0                              

    D)  \[\frac{1}{3}\]

    Correct Answer: B

    Solution :

    Let F is spring force and \[F=kx\] In position D, spring in equilibrium \[N+mg\,-F=0\]              ? In bottom most position. \[N+F-mg=\frac{m{{v}^{2}}}{r}\](since spring is elongated) \[x-\frac{3R}{4}-\frac{R}{2}\,=\frac{R}{4}\] in position D and \[x=R-\frac{3R}{4}\,=\frac{R}{4}\] in bottom position B hence F is same in D and B position. Now from  and  \[N+(N+mg)-mg=\frac{m{{v}^{2}}}{r}\]\[2N=\frac{m{{v}^{2}}}{r}\]\[\frac{N}{\frac{m{{v}^{2}}}{r}}=\frac{1}{2}\]


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