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 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 $I_{B B}=m_A l^2$
$\frac{I_{B B}}{I_{A A}}=3$ (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))