question_answer

The centre of mass of triangle shown in figure has coordinates

A)  \[x=\frac{h}{2},y=\frac{b}{2}\]           

B)  \[x=\frac{b}{2},y=\frac{h}{2}\]

C)  \[x=\frac{b}{3},y=\frac{h}{3}\]         

D)         \[x=\frac{h}{3},y=\frac{b}{3}\]

Correct Answer: C

Solution :

 We can assume that three particles of equal mass m placed at the corners of triangle \[\overset{\to }{\mathop{{{r}_{1}}}}\,=0\widehat{i}+0\widehat{j,}\overrightarrow{{{r}_{2}}}=b\widehat{i}+0\widehat{j}\]and \[\overset{\to }{\mathop{{{r}_{3}}}}\,=0\widehat{i}+h\widehat{j}\] \[\therefore {{\widehat{r}}_{cm}}=\frac{{{m}_{1}}{{\overrightarrow{r}}_{1}}+{{m}_{2}}{{\overrightarrow{r}}_{2}}+{{m}_{3}}{{\overrightarrow{r}}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}=\frac{b}{3}\widehat{i}+\frac{h}{3}\widehat{j}\] i.e. coordinates of centre of mass is \[\left( \frac{b}{3},\frac{h}{3} \right)\]