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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    A uniform rod AB is suspended from a point X, at a variable distance \[x\] from \[A\], as shown. To make the rod horizontal, a mass \[m\] is suspended from its end\[A\]. A set of \[(m,x)\] values is recorded. The appropriate variables that give a straight line, when plotted, are:                [JEE Online 15-04-2018]

    A) \[m,\frac{l}{x}\]                     

    B) \[m,\frac{l}{{{x}^{2}}}\]               

    C) \[m,x\]             

    D) \[m,{{x}^{2}}\]

    Correct Answer: A

    Solution :

    Let, 'l' and 'M' be the length and mass of rod for equilibrium, taking moment about X, we get \[m\times x=M\times (\frac{l}{2}-x)\] this can be written as, \[m=(\frac{Ml}{2})\frac{l}{x}-M\] comparing with straight line equation, \[y=mx+c\] we can say that, \[m\propto \frac{l}{x}\]


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