The position vector of the centre of mass $\vec r\, cm$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

820-1093

  • [JEE MAIN 2019]
  • A

    $\vec r\,cm = \frac{{13}}{8}L\hat x + \frac{5}{8}L\hat y$

  • B

    $\vec r\,cm = \frac{{5}}{8}L\hat x + \frac{13}{8}L\hat y$

  • C

    $\vec r\,cm = \frac{{3}}{8}L\hat x + \frac{11}{8}L\hat y$

  • D

    $\vec r\,cm = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y$

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