Question

A square of side 4 cm and uniform thickness is divided into four squares. The square portion $A^{{}^{\prime }}A{B}^{{}^{\prime }}D$ is removed and the removed portion is placed over the square $B^{\prime }B{C}^{\prime }D$. The new position of centre of mass is-

• A. $($ 2 cm, 2 cm$)$
• B. $($2 cm, 3 cm$)$
• C. $($2 cm, 2.5 cm$)$
• D. $($3 cm, 3 cm$)$

Answer 1

The correct option is C $($2 cm, 2.5 cm$)$

The $CM$ of each of the smaller square is at the centre of the smaller square.
Co-ordinates of the $CM$ of the smaller squares are as follows:
square $O{A}^{\prime }D{D}^{{}^{\prime }}:(x=1,y=1)$
square $C{C}^{{}^{\prime }}D{D}^{{}^{\prime }}:(x=1,y=3)$
square $C^{{}^{\prime }}B{B}^{{}^{\prime }}D:(x=3,y=3)$
Since square $A^{{}^{\prime }}A{B}^{{}^{\prime }}D$ is removed and put on top of square $C^{{}^{\prime }}B{B}^{{}^{\prime }}D$, we need to count two squares at $C^{{}^{\prime }}B{B}^{{}^{\prime }}D$ i.e the area is to be taken two times the area of a smaller square.

$X_{CM}=\frac{A_1{x}_1+A_2{x}_2+A_3{x}_3+A_4{x}_4}{A_1+A_2+A_3+A_4}$
$Y_{CM}=\frac{A_1{y}_1+A_2{y}_2+A_3{y}_3+A_4{y}_4}{A_1+A_2+A_3+A_4}$;

$A_1=A_2=A_3=A_4=4$;
$y_1=y_2=1$;
$x_3=x_4=3$;
$y_1=1$;
$y_2=y_3=y_4=3$;

$X_{CM}=\frac{4\times 1+4\times 1+4\times 3+4\times 3}{4+4+4+4}=2$ cm
$Y_{CM}=\frac{4\times 1+4\times 3+4\times 3+4\times 3}{4+4+4+4}=2.5$ cm