Consider an elliptic disc having unifrom distribution of mass $m$ as shown in figure. The moment of inertia (M.I.) about the major and minor axis are $\frac{1}{4} m b^{2} and \frac{1}{4} m a^{2}$ respectively. Then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The M.I. about a chord of length 2r passing through O and lying in plane of elliptic disc is $\frac{1}{4} \frac{m a^{2} b^{2}}{r^{2}}$

The M.I. about a line perpendicular to the plane of elliptic disc and passing through the centre O is $\frac{1}{4} m (a^{2} + b^{2})$

The M.I. about a line, in its plane, passing through its centre and inclined at an angle $θ$ to the major axis is $\frac{1}{4} m (b^{2} \cos^{2} θ + a^{2} \sin^{2} θ)$

The M.I. about a line, in its plane, passing through its centre and inclined at an angle $θ$ to the minor axis is $\frac{1}{4} m (b^{2} \sin^{2} θ + a^{2} \cos^{2} θ)$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We know that
$\begin{array}{l} I_{P Q} = I_{x} \cos^{2} θ + I_{y} \sin^{2} θ - I_{x y} \sin 2 θ \\ I_{P Q} = \frac{1}{4} m b^{2} \cos^{2} θ + \frac{1}{4} m a^{2} \sin^{2} θ [∵ I_{x y} = 0] \end{array}$
The equation of ellipse is
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\begin{matrix} \frac{r^{2}}{a^{2}} \cos^{2} θ + \frac{r^{2}}{b^{2}} \sin^{2} θ = 1 \\ a^{2} \sin^{2} θ + b^{2} \cos^{2} θ = \frac{a^{2} b^{2}}{r^{2}} \end{matrix}$