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,
$Y_{CM} = \frac{m_1y_1 + m_2 y_2}{m_1+ m_2}$
$\Rightarrow 0 = m_1 y_1 + m_2y_2$
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} $
$=\frac{3}{4}$
$\therefore m_{1} :m_{2}=3:4$