Question

Masses 8,2,4,2 kg are placed at the corners A,B,C,D respectively of a square ABCD of diagonal 80 cm. The distance of centre of mass from A will be

• A. 20 cm
• B. 30 cm
• C. 40 cm
• D. 60 cm

Answer 1

The correct option is B 30 cm

Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner (given in problem) Diagonal of square $d=a\sqrt{2}=80cm\Rightarrow a=40\sqrt{2}cm$
$m_1=8kg,m_2=2kg,m_3=4kg,m_4=2kg$
Let $\overrightarrow{r_1},\overrightarrow{r_2},\overrightarrow{r_3},\overrightarrow{r_4}$ are the position vectors of respective masses
$\overrightarrow{r_1}=0\hat{i}+0\hat{j},\overrightarrow{r_2}=a\hat{i}+0\hat{j},\overrightarrow{r_3}=a\hat{i}+a\hat{j},\overrightarrow{r_4}=0\hat{i}+a\hat{j}$
From the formula of centre of mass
$\overrightarrow{r}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}+m_4\overrightarrow{r_4}}{m_1+m_2+m_3+m_4}=15\sqrt{2}i+15\sqrt{2}j$
$\therefore$ co-ordinates of centre of mass $(15\sqrt{2},15\sqrt{2})$ and co-ordination of the corner = (0,0)
From the formula of distance between two points $(x_1,y_1)$ and $(x_2,y_2)$
distance = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(15\sqrt{2}-0{)}^2+(15\sqrt{2}-0{)}^2}=\sqrt{900}=30cm$