Q126 / 150 · MHT CET 2020 (20 Oct Shift 1) · Physics
Two discs having moment of inertia $\mathrm{l}_{1}$ and $\mathrm{I}_{2}$ are made from same material have
same mass. Their thickness and radii are $\mathrm{t}_{1}, \mathrm{t}_{2}$ and $\mathrm{R}_{1}, \mathrm{R}_{2}$ respectively. The relation
between moment of inertia of each disc about an axis passing through its centre
and perpendicular to its plane and its thickness is
- Correct answerA$\mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}$
- B$\mathrm{I}_{1} \mathrm{t}_{2}^{2}=\mathrm{I}_{2} \mathrm{t}_{1}^{2}$
- C$\mathrm{I}_{1} \mathrm{t}_{2}=\mathrm{I}_{2} \mathrm{t}_{1}$
- D$\mathrm{I}_{1} \mathrm{t}_{1}^{2}=\mathrm{I}_{2} \mathrm{t}_{2}^{2}$
Solution
$I_{1}=\frac{m R_{1}^{2}}{2}$
$I_{2}=\frac{m R_{2}^{2}}{2}$
$\frac{I_{1}}{I_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{t_{2}}{t_{1}}$
$\therefore \mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}$
$\mathrm{m}_{1}=\pi \mathrm{R}_{1}^{2} \mathrm{t}_{1}$
$\mathrm{~m}_{2}=\pi \mathrm{R}_{2}^{2} \mathrm{t}_{2}$\
$\mathrm{m}_{1}=\mathrm{m}_{2}$
$\therefore \mathrm{R}_{1}^{2} \mathrm{t}_{1}=\mathrm{R}_{2}^{2} \mathrm{t}_{2}$
Hence, the correct answer is option: 'A'.