Question

A uniform rod of length $l=100cm$ is bent at its mid-point to make ${90}^{\circ }$ angle. The distance of the centre of mass from the centre of the rod is

• A. 36.1 cm
• B. 25.2 cm
• C. 17.7 cm
• D. Zero

Answer 1

The correct option is C

17.7 cm

Let the rod POQ of length $l$ be bent at its mid-point O so that $\angle POQ={90}^{\circ }$ as shown in the figure. The mass of part PO of length $\frac{l}{2}$ can be taken to be concentrated at its mid-point A whose co-ordinates are $(0,\frac{l}{4})$ and of part OQ of length $\frac{l}{2}$ at its mid-point B whose coordinates are $(\frac{l}{4},0)$. The centre of mass of these two equal masses is at mid-point C between A and B. The coordinates of C are $\left(\frac{l}{8}\right),\left(\frac{l}{8}\right)$.

$\therefore OC=\sqrt{(OE{)}^2+(CE{)}^2}=\sqrt{{\left(\frac{l}{8}\right)}^2+{\left(\frac{l}{8}\right)}^2}=\frac{l}{\sqrt{32}}=\frac{100cm}{\sqrt{32}}=17.7cm$
Hence, the correct choice is (c).