Question

$A B$ and $C D$ are two identical rods each of length $l$ and mass $m$ joined to form a cross. The moment of inertia of these two rods about a bisector $(X Y)$ of angle between the rods is

• A. $\frac{m l^{2}}{12}$
• B. $\frac{m l^{2}}{3}$
• C. $\frac{2 m l^{2}}{3}$
• D. $\frac{m l^{2}}{6}$

Answer 1

Answer: (A)

Moment of inertia of rods $A B$ and $C D$ about an axis
passing through their point of intersection $O$ and perpendícular
to the plane of the rods is
$I_{O}=\frac{m l^{2}}{12}+\frac{m l^{2}}{12}=\frac{m l^{2}}{6}$

In figure, $X Y$ and $X^{\prime} Y$ are 'two
mutually perpendicular axes passing
through $O$ lying in the plane of the two
rods. By symmetry
$I_{X Y}=I_{X^{\prime} Y^{\prime}}=I,$ say
From the theorem of perpendicular axes
$I_{X Y}+I_{X^{\prime} Y^{\prime}}=I_{O} $
$\text { or } 2 I=\frac{m l^{2}}{6} \,\,\, \text { or } \,\,\, I=\frac{m l^{2}}{12} $