Question

Figure shows a square plate of uniform thickness and side length $\sqrt{2}$m. One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is

• A. 1/3m
• B. 1/2 m
• C. 1/6 m
• D. 1/8m

Answer 1

The correct option is

C

1/6 m

As we cut a part, from 1st quadrant, the COM will shift to third quadrant due to its symmetry, lets mark the COM of each segment as shown in figure,

according to the question

$2L=\sqrt{2}⟹L=\frac{1}{\sqrt{2}}$

$X_{com}=\frac{(m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4)}{m_1+m_2+m_3+m_4}$

$=\frac{(0\left(\frac{L}{2}\right)+M(-\frac{L}{2})+M(-\frac{L}{2})+M(+\frac{L}{2}))}{M+M+M}=-\frac{ML}{6M}=-\frac{L}{6}$

$Y_{com}=\frac{(m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4)}{m_1+m_2+m_3+m_4}$

$=\frac{(0\left(\frac{L}{2}\right)+M(+\frac{L}{2})+M(-\frac{L}{2})+M(-\frac{L}{2}))}{M+M+M}=-\frac{ML}{6M}=-\frac{L}{6}$

$X_{com},Y_{com}=(-\frac{L}{6},-\frac{L}{6})$

$distance=\sqrt{{\left(\frac{L}{6}\right)}^2+{\left(\frac{L}{6}\right)}^2}=\sqrt{\frac{2L^2}{36}}=\frac{L}{6}\sqrt{2}=\frac{1}{6}m$