question_answer

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-                       [JEE Main 12-Jan-2019 Morning]

A)   \[\vec{r}\,cm=\frac{11}{8}L\hat{x}+\frac{3}{8}L\,\hat{y}\]

B)        \[\vec{r}\,cm=\frac{13}{8}L\hat{x}+\frac{5}{8}L\,\hat{y}\]

C)   \[\vec{r}\,cm=\frac{3}{8}L\hat{x}+\frac{11}{8}L\,\hat{y}\]

D)   \[\vec{r}\,cm=\frac{5}{8}L\hat{x}+\frac{13}{8}L\,\hat{y}\]

Correct Answer: B

Solution :

[b] Letsoand