MCQ
A thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma (r) = kr^2$ where $r$ is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is
  • A
    $\frac{{M{R^2}}}{2}$
  • B
    $\frac{{M{R^2}}}{3}$
  • C
    $\frac{{M{R^2}}}{6}$
  • $\frac{{2M{R^2}}}{3}$

Answer

Correct option: D.
$\frac{{2M{R^2}}}{3}$
d
${I_{Disc}} = \int\limits_0^R {\left( {dm} \right){r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right){r^2}} } $

${I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}\,\,\,\,\,\,\,\,\,Mass\,of\,disc$

${I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M = } \int\limits_0^R {2\pi rdr\,k{r^2}} $

${I_{Disc}} = 2\pi k\left( {\frac{{{r^6}}}{6}} \right)_0^R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M = 2\pi k\int\limits_0^R {{r^3}dr} $

${I_{Disc}} = 2\pi k\frac{{{R^6}}}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M = 2\pi k\left. {\frac{{{r^4}}}{4}} \right|_0^R$

${I_{Disc}} = \frac{{\pi k{R^6}}}{3} = \left( {\frac{{\pi k{R^4}}}{2}} \right)\frac{{{R^2}2}}{3}\,M = 2\pi k\left. {\frac{{{r^4}}}{4}} \right|_0^R$

${I_{Disc}} = \frac{{M2{R^2}}}{3}\,\,\,\,\,;\,\,\,\,\,\,\,\,{I_{Disc}} = \frac{2}{3}M{R^2}$

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