Question

A uniform metal rod of length 1 m is bent at ${90}^0$, so as to form two arms of equal length. Then centre of mass of this bent rod is

• A. On the bisector of the angle , $\left(\frac{1}{\sqrt{2}}\right)$ m from vertex
• B. On the bisector of angle. $\left(\frac{1}{2\sqrt{2}}\right)$m from vertex
• C. On the vertex
• D. On any end point of the side other than vertex

Answer 1

The correct option is A On the bisector of the angle , $\left(\frac{1}{\sqrt{2}}\right)$ m from vertex

After bending the $1m$ rod, the arms will have length -$\frac{1}{2}m$ each

consider $0$ as origin $(0,0)$

So, centre of man of rod $OA$ is at $(0,\frac{1}{4})$ and that of $B$ is at $(\frac{1}{4},0)$

From symmetry (man of $OA=OB$) location of C is $C\equiv (\frac{1}{4},\frac{1}{4})$

Thus length $OC=\sqrt{{\left(\frac{1}{4}\right)}^2+{\left(\frac{1}{4}\right)}^2}=\frac{1}{\sqrt{2}}$