JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    A rod of mass M which is non-uniformly distributed over its length L has been considered. Its linear mass density variation with distance x from left end is shown in the figure. From this information, we can conclude that the centre of mass of rod                                                         

    A)  is on left of its mid-point                    

    B)  is on right of its mid-point

    C)  is coinciding with its mid-point             

    D)  information is insufficient

    Correct Answer: A

    Solution :

     Idea The given graph is between mass per unit length versus length of rod. So, its area will give mass of rod and by analyzing mass distribution we can observe the position of centre of mass. Area under\[\lambda \]versus x curve gives mass. For any x and y area of I > area of II, it means more mass as on the left side, so centre of mass would be on left half. TEST Edge In this question linear mass density varies with\[\lambda =kx\]. (k = positive constant) Let the mass of small area = dm, So, \[dm=\lambda dx=kxdx\] \[\Rightarrow \]So\[{{X}_{CM}}=\int_{0}^{L}{x}\,dm=\int_{0}^{L}{k{{x}^{2}}}dx\] \[=k\left( \frac{{{x}^{3}}}{3} \right)_{0}^{L}=\frac{k{{L}^{3}}}{3}\]

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