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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2012 · Paper 1 · Physics · Rotational Motion

A lamina is made by removing a small disc of diameter 2 R from a...

A lamina is made by removing a small disc of diameter 2R2 R from a bigger disc of uniform mass density and radius 2R2 R, as shown in the figure. The moment of inertia of this lamina about axes passing though OO and PP is IOI_{O} and IPI_{P} respectively. Both these axes are perpendicular to the plane of the lamina. The ratio IP/IOI_{P} / I_{O} to the nearest integer is A lamina is made by removing a small disc of diameter 2 R from a bigger disc of uniform mass density and radius 2 R ,

Correct answer

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Step-by-step solution

Let σ\sigma be the surface mass density. Moment of inertia of the lamina about axes passing through 'O' IO=12σ[π(2R)2]×(2R)2−=132πσR4[12(σπR2)2+σ(πR2)×R2]\begin{array}{l} I_{O}=\frac{1}{2} \sigma\left[\pi(2 R)^{2}\right] \times(2 R)^{2}- \\ =\frac{13}{2} \pi \sigma R^{4} \quad\left[\frac{1}{2}\left(\sigma \pi R^{2}\right)^{2}+\sigma\left(\pi R^{2}\right) \times R^{2}\right] \end{array} Moment of inertia of the lamina about axes passing through 'P' IP=8πσR4+σπ(2R)2×(2R)2[12σ(πR2)R2+σ(πR2)((2R)2+R2)2]\begin{array}{l} I_{P}=8 \pi \sigma \mathrm{R}^{4}+\sigma \pi(2 \mathrm{R})^{2} \times(2 \mathrm{R})^{2} \\ {\left[\frac{1}{2} \sigma\left(\pi R^{2}\right) R^{2}+\sigma\left(\pi R^{2}\right)\left(\sqrt{(2 R)^{2}+R^{2}}\right)^{2}\right]} \end{array} =24πσR4−5.5σπR4=18.5πσR4∴IPIO=18.5πσR4132πσR4=3713≈3\begin{array}{l} =24 \pi \sigma \mathrm{R}^{4}-5.5 \sigma \pi \mathrm{R}^{4}=18.5 \pi \sigma \mathrm{R}^{4} \\ \therefore \quad \frac{I_{P}}{I_{O}}=\frac{18.5 \pi \sigma R^{4}}{\frac{13}{2} \pi \sigma R^{4}}=\frac{37}{13} \approx 3 \end{array}
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