Question bank/JEE Main/Physics

A thin walled hollow cylinder is closed from both the ends. The
mass of the curved part is m and the mass of each circular part is also
m. The radius of the cylinder is R and its length is 2R. What will be
the moment of inertia of this hollow cylinder about the axis,
perpendicular to its length and passing through its centre of mass?

A$\frac{10}{3}mR^{2}$B$\frac{5}{3}mR^{2}$C$\frac{5}{6}mR^{2}$D$\frac{11}{3}mR^{2}$

Show answer and solution▾

First find I of the hollow tube
$I_{1} = \int\frac{{dmR}^{2}}{2} + \int_{x = - \frac{\mathcal{l}}{2}}^{x = + \frac{\mathcal{l}}{2}}{(\frac{m}{\mathcal{l}}dx)}x^{2}$
$I_{1} = \frac{{mR}^{2}}{2} + \frac{m^{\mathcal{l}^{2}}}{12}$ where
$\mathcal{l} = 2R$
So $I_{1} = \frac{5}{6}{mR}^{2}$
Now moment of inertias of each circular plate about that axis.
$I = \frac{mR^{2}}{4} + mR^{2} = \frac{5}{4}mR^{2}$
$I_{total} = \frac{5}{6}{mR}^{2} + 2 \times \left( \frac{5}{4}{mR}^{2} \right)$
$I_{total} = \frac{10}{3}{mR}^{2}$

Source: Grass-G / P-20 Physcis Paper 12 Dec.docx