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 Online Paper Held On 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 :

\[{{\vec{r}}_{1}}=L(\hat{x}+\hat{y})\] \[{{\vec{r}}_{2}}=2L\hat{x}+\frac{L}{2}\hat{y}\] \[{{\vec{r}}_{3}}=2.5L\hat{x}\] \[{{\vec{r}}_{cm}}=\frac{{{m}_{1}}{{{\vec{r}}}_{1}}+{{m}_{2}}{{{\vec{r}}}_{2}}+{{m}_{3}}{{{\vec{r}}}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] Let\[{{m}_{2}}=m,\]so\[{{m}_{3}}=m\]and\[{{m}_{1}}=2m\] \[\therefore \]\[{{\vec{r}}_{cm}}=\frac{1}{4}\left[ 2L(\hat{x}+\hat{y})+\left( 2L\hat{x}+\frac{L}{2}\hat{y} \right)+2.5L\hat{x} \right]\] \[=\frac{1}{4}(6.5\hat{x}+2.5\hat{y})=\frac{13L}{8}\hat{x}+\frac{5L}{8}\hat{y}.\]