Question

Three rectangular plates $A,B$ and $C$ made up of three different materials are joined together to form a large rectangular plate as shown in figure. If the densities of $A,B$ and $C$ are in the ratio $2:5:3$, then choose the correct option.

• A. $COM$ lies inside $A$
• B. $COM$ lies inside $B$
• C. $COM$ lies inside $C$
• D. $COM$ lies outside all three sheets

Answer 1

The correct option is C $COM$ lies inside $C$

Let us consider plate in $x,y$ plane, taking origin at $\text{O}$


And we know that COM of a rectangular sheet of length $l$ and width $b$ is at $(\frac{l}{2},\frac{b}{2})$
Therefore,
COM of sheet $A$ is $(2,2.5)$
COM of sheet $B$ is $(7,4)$
COM of sheet $C$ is $(7,1.5)$

Area of sheet $A$ is $A_1=5\times 4=20\text{sq units}$
Area of sheet $B$ is $A_2=6\times 2=12\text{sq units}$
Area of sheet $C$ is $A_3=6\times 3=18\text{sq units}$

Let density of $A,B$ and $C$ be $2\sigma ,5\sigma ,3\sigma$
Mass of sheet $A=$ area $\times$ density $=20\times 2\sigma =40\sigma$
Mass of sheet $B=$ area $\times$ density $=12\times 5\sigma =60\sigma$
Mass of sheet $C=$ area $\times$ density $=18\times 3\sigma =54\sigma$

$X_{COM}$ of combined sheet is given by
$=\frac{x_1{m}_1+x_2{m}_2+x_3{m}_3}{m_1+m_2+m_3}$
$=\frac{2\times 40\sigma +7\times 60\sigma +7\times 54\sigma }{40\sigma +60\sigma +54\sigma }$
$=5.7$
$Y_{COM}$ of combined sheet $=\frac{y_1{m}_1+y_2{m}_2+y_3{m}_3}{m_1+m_2+m_3}$
$=\frac{2.5\times 40\sigma +4\times 60\sigma +1.5\times 54\sigma }{40\sigma +60\sigma +54\sigma }$
$=2.73$
Therefore COM lies at $(5.7,2.73)$ i.e inside sheet $C$.