JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- MCQ
Mass per unit area of a circular disc of radius \(\mathrm{a}\) depends on the distance \(\mathrm{r}\) from its centre as \(\sigma(\mathrm{r})=\mathrm{A}+\mathrm{Br} .\) The moment of inertia of the disc about the the axis, perpendicular to the plane and passing through its centre is
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Answer & Solution
Correct Answer
(A) \(2 \pi \mathrm{a}^{4}\left(\frac{\mathrm{A}}{4}+\frac{\mathrm{aB}}{5}\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{dI}=\mathrm{dm}^{2}\)
\(\mathrm{dI}=\sigma 2 \pi \mathrm{r}\) dr \(\mathrm{r}^{2}\)
\(\mathrm{dI}=2 \pi(\mathrm{A}+\mathrm{Br}) \mathrm{r}^{3} \mathrm{dr}\)
\(\int \mathrm{dI}=2 \pi \int_{0}^{\mathrm{a}}\left(\mathrm{Ar}^{3}+\mathrm{Br}^{4}\right) \mathrm{dr}\)
\(I=2 \pi a^{4}\left(\frac{A}{4}+\frac{B a}{5}\right)\)
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