Q72 / 90 · KVPY 2015 (SB/SX) · Physics
The moments of inertia of a non-uniform circular disc (of mass $\mathrm{M}$ and radius $\mathrm{R}$ ) about four mutually perpendicular tangents $\mathrm{AB}, \mathrm{BC} \mathrm{CD}, \mathrm{DA}$ are $I_{1}, I_{2}, I_{3}$ and $I_{4}$ respectively (the square ABCD circumscribes the circle.) The distance of the center of mass of the disc from its geometrical center is given by.
- Correct answerA$\frac{1}{4 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$
- B$\frac{1}{12 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$
- C$\frac{1}{3 M R} \sqrt{\left(I_{1}-I_{2}\right)^{2}+\left(I_{3}-I_{4}\right)^{2}}$
- D$\frac{1}{2 M R} \sqrt{\left(I_{1}+I_{3}\right)^{2}+\left(I_{2}+I_{4}\right)^{2}}$
Solution
$$
\begin{array}{l}
I_{1}=I_{c m}+M(R-y)^{2} \\
I_{3}=I_{c m}+M(R+y)^{2}
\end{array}
$$
So $I_{3}-I_{1}=M\left[(R+y)^{2}-(R-y)^{2}\right\rfloor$
same as $I_{4}-I_{2}=M\left[(R+x)^{2}-(R-x)^{2}\right]$
Ву
Distance of $\mathrm{CM}$ from center of disc $=\sqrt{x^{2}+y^{2}}=\frac{1}{4 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$
Hence, the correct answer is option: 'A'.