Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ r = A…

1. $2\pi a^4\left(\frac{A}{4}+\frac{aB}{5}\right)$
2. $2\pi a^4\left(\frac{aA}{4}+\frac{B}{5}\right)$
3. $\pi a^4\left(\frac{A}{4}+\frac{aB}{5}\right)$
4. $2\pi {\text{a}}^4\left(\frac{\text{A}}{4}+\frac{\text{B}}{5}\right)$

Solution

$\mathrm{\sigma }=\mathrm{A}+\mathrm{B}\mathrm{r}$
$\int \mathrm{d}\mathrm{m}=\int \left(\mathrm{A}+\mathrm{B}\mathrm{r}\right)2\mathrm{\pi }\mathrm{r}\mathrm{d}\mathrm{r}\mathrm{}\mathrm{}$
$\mathrm{I}=\int \mathrm{d}\mathrm{m}\mathrm{}{\mathrm{r}}^2$
$\underset{0}{\overset{a}{\int }}\left(A+Br\right)2\pi r^3\mathrm{d}\mathrm{r}$
$=2\pi \left(\mathrm{A}\frac{{\mathrm{a}}^4}{4}+\mathrm{B}\frac{{\mathrm{a}}^5}{5}\right)$
$=2\pi a^4\left(\frac{A}{4}+\frac{aB}{5}\right)$

Asked in: JEE Main 2020 (07 Jan Shift 2)

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