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 :
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