Mass per unit area of a circular disc of radius a depends on the distance r from its centre as \sigma ( r ) = A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

2\pi a^4( A \over 4 + aB \over 5 )

\pi a^4( A \over 4 + aB \over 5 )

2\pi a^4( aA \over 4 + B \over 5 )

2\pi a^4( A \over 4 + B \over 5 )

Rotational Motion PYQ — JEE Main 2020 · 7 Jan · Shift 2 · Q57 with Solution

View written solutionFree

Correct answer: A

Given surface mass density

$\sigma(r)=A+Br$

We need the moment of inertia of the disc about the axis perpendicular to the plane through the centre.
Take a thin ring element

Consider a ring of radius $r$ and thickness $dr$ .

Its area is $dS=2\pi r\,dr$

Hence its mass is $dm=\sigma(r)\,dS=(A+Br)(2\pi r\,dr)$
Moment of inertia of the ring element

For a ring about its central perpendicular axis, $dI=r^2\,dm$

Therefore, $dI=r^2(A+Br)(2\pi r\,dr)$ $dI=2\pi (A+Br)r^3\,dr$
Integrate from $r=0$ to $r=a$

$I=\int_0^a dI=2\pi\int_0^a (A+Br)r^3\,dr$

Expand: $I=2\pi\int_0^a (Ar^3+Br^4)\,dr$

Now integrate: $I=2\pi\left[A\frac{r^4}{4}+B\frac{r^5}{5}\right]_0^a$

$I=2\pi\left(\frac{Aa^4}{4}+\frac{Ba^5}{5}\right)$
Factor out $a^4$

$I=2\pi a^4\left(\frac{A}{4}+\frac{aB}{5}\right)$
Compare with options

This matches Option A: $2\pi a^4\left(\frac{A}{4}+\frac{aB}{5}\right)$

Final Answer: Option A