Q187 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
A thin uniform rod of mass 'M' and length 'L' lies in a $x$-$y$ plane. Its centre of origin is at $(0,0)$ and is oriented along the $x$-axis. What is the moment of inertia of this rod about an axis passing through point $( 0, 2L/3)$ parallel to $z$- axis and perpendicular to the plane.
- A$\dfrac{36}{19} M L^2$
- B$\dfrac{2}{3} M L^2$
- Correct answerC$\dfrac{19}{36} M L^2$
- D$\dfrac{3}{2} M L^2$
Solution
The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its centre of mass and perpendicular to its length is $I_{cm} = \dfrac{M L^2}{12}$.
The new axis passes through the point $(0, \dfrac{2L}{3})$ and is parallel to the $z$-axis. The perpendicular distance $d$ between the parallel axes is $d = \dfrac{2L}{3}$.
Using the parallel axis theorem, the moment of inertia about the new axis is given by:
$I = I_{cm} + M d^2$
$I = \dfrac{M L^2}{12} + M \left(\dfrac{2L}{3}\right)^2$
$I = \dfrac{M L^2}{12} + \dfrac{4 M L^2}{9}$
$I = \dfrac{3 M L^2 + 16 M L^2}{36}$
$I = \dfrac{19}{36} M L^2$
Answer: $\dfrac{19}{36} M L^2$
The new axis passes through the point $(0, \dfrac{2L}{3})$ and is parallel to the $z$-axis. The perpendicular distance $d$ between the parallel axes is $d = \dfrac{2L}{3}$.
Using the parallel axis theorem, the moment of inertia about the new axis is given by:
$I = I_{cm} + M d^2$
$I = \dfrac{M L^2}{12} + M \left(\dfrac{2L}{3}\right)^2$
$I = \dfrac{M L^2}{12} + \dfrac{4 M L^2}{9}$
$I = \dfrac{3 M L^2 + 16 M L^2}{36}$
$I = \dfrac{19}{36} M L^2$
Answer: $\dfrac{19}{36} M L^2$
Hence, the correct answer is option: 'C'.