QJEEM19Pha9d4: JEE Main 2019 Physics - Class 11 JEEM19Pha9d4 Solution: Let the moment of inertia of a hollow cylinder of length 30 cm…

Consider an element of radius x and thickness dx

Mass of element, dm = $\sigma 2\pi x\left( {dx} \right)$

Here, $\sigma$ = mass per unit area = ${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$

Moment of inertia of element, dI = (dm)x²

$\Rightarrow$ I = $\sigma 2\pi \int\limits_r^R {{x^3}dx}$

= $\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$

= ${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$

= ${m \over 2}\left( {{R^2} + {r^2}} \right)$ .....(i)

Moment of inertia of thin cylinder of same mass,

I = m $r_0^2$ ......(ii)

$\Rightarrow$ m $r_0^2$ = ${m \over 2}\left( {{R^2} + {r^2}} \right)$

$\Rightarrow$ $r_0^2$ = 250

$\Rightarrow$ r₀ $\simeq$ 16 cm