JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(\frac{{M{R^2}}}{6}\)
- D \(\frac{{2M{R^2}}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{{2M{R^2}}}{3}\)
Step-by-step Solution
Detailed explanation
\({I_{Disc}} = \int\limits_0^R {\left( {dm} \right){r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right){r^2}} } \) \({I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}\,\,\,\,\,\,\,\,\,Mass\,of\,disc\)…
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