Question

A point mass $m_{A}$ is connected to a point mass $m_{B}$ by a massless rod of length $l$ as shown in the figure. It is observed that the ratio of the moment of inertia of the system about the two axes $B B$ and $A A$, which are parallel to each other and perpendicular to the rod is $\frac{I_{B B}}{I_{A A}}=3$
The distance of the centre of mass of the system from the $\operatorname{mass} A$ is

• A. $\frac{1}{2}l$
• B. $\frac{1}{4}l$
• C. $\frac{2}{3} l$
• D. $\frac{3}{4} l$

Answer 1

Answer: (B)

$I_{A A}=m_{B} l^{2}$ and $I_{B B}=m_{A} l^{2}$
$\frac{I_{B B}}{I_{A A}}=3 \quad($ Given $) ; \therefore \frac{m_{A}}{m_{B}}=3 \,\,\,\,\,\, ...(i)$
Let $x$ be the distance of centre of mass from mass $A$.
$\therefore m_{A} x=m_{B}(l-x)$ or $\frac{m_{A}}{m_{B}} x=l-x$
or $3 x=l-x$ or $x=\frac{l}{4} \,\,\,\,\,$ (Using(i))