Question

Two uniform plates of the same thickness and are abut of different materials, one shaped like an isosceles triangle and the other shaped like a rectangle are joined together to form a composite body as shown in the figure alongside. If the centre of mass of the composite body is located at the mid-point of their common side, then the ratio between masses of the triangle to that of the rectangle is

• A. $1:1$
• B. $4:3$
• C. $3:4$
• D. $2:1$

Answer 1

Answer: (C)

Let given plate has dimensions, as shown below.

Given, area of triangle = area of rectangle
$\Rightarrow \frac{1}{2}a\times h=ab$
$\Rightarrow \frac{h}{2}=b$
$\Rightarrow \frac{b}{h}=\frac{1}{2}$
From standard results, centre of mass of triangular part is $\frac{h}{3}$ above origin chosen, which is a t centre of mass of complete (lamina).
Also, centre of mass of rectangular part is $\frac{b}{2}$ below the origin (shown in figure).
Now, for complete lamina,
$Y_{CM}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$
$\Rightarrow 0=m_1{y}_1+m_2{y}_2$
Here, $m_1=$ mass of triangular portion,
$m_2=$ mass of rectangular portion,
$y_1=$ position of centre of mass of triangular portion $=\frac{h}{3}$
and $y_2=$ position of centre of mass of $=-\frac{b}{2}$.
$\Rightarrow 0=m_1\frac{h}{3}+m_2\left(-\frac{b}{2}\right)$
$\Rightarrow \frac{m_1}{m_1}=\frac{3}{2}\times \frac{b}{h}$
$=\frac{3}{2}\times \frac{1}{2}$
$=\frac{3}{4}$
$\therefore m_1:m_2=3:4$