Two discs having moment of inertia $\mathrm{l}_{1}$ and $\mathrm{I}_{2}$ are made from same material have…

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
  1. $\mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}$
  2. $\mathrm{I}_{1} \mathrm{t}_{2}^{2}=\mathrm{I}_{2} \mathrm{t}_{1}^{2}$
  3. $\mathrm{I}_{1} \mathrm{t}_{2}=\mathrm{I}_{2} \mathrm{t}_{1}$
  4. $\mathrm{I}_{1} \mathrm{t}_{1}^{2}=\mathrm{I}_{2} \mathrm{t}_{2}^{2}$

Solution

$I_{1}=\frac{m R_{1}^{2}}{2}$ $I_{2}=\frac{m R_{2}^{2}}{2}$ $\frac{I_{1}}{I_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{t_{2}}{t_{1}}$ $\therefore \mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}$ $\mathrm{m}_{1}=\pi \mathrm{R}_{1}^{2} \mathrm{t}_{1}$ $\mathrm{~m}_{2}=\pi \mathrm{R}_{2}^{2} \mathrm{t}_{2}$\ $\mathrm{m}_{1}=\mathrm{m}_{2}$ $\therefore \mathrm{R}_{1}^{2} \mathrm{t}_{1}=\mathrm{R}_{2}^{2} \mathrm{t}_{2}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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