QJEEAdv12Phb5ef: JEE Advanced 2012 Physics - Class 11 JEEAdv12Phb5ef Solution: A lamina is made by removing a small disc of diameter 2R from a …

Then, ${m_T} = \pi {(2R)^2}\sigma = 4\pi {R^2}\sigma$

${m_C} = \pi {(R)^2}\sigma = \pi {R^2}\sigma$

For I_P

${I_R} = {I_T} - {I_C}$

$= {3 \over 2}{m_T}{(2R)^2} - \left[ {{1 \over 2}{m_C}{R^2} + {m_C}{r^2}} \right]$

$= {3 \over 2}(4\pi {R^2}\sigma )(4{R^2}) - \left[ {{1 \over 2}(\pi {R^2}\sigma ) + (\pi {R^2}\sigma )(5{R^2})} \right]$

$= (18.5\pi {R^4}\sigma )$

For I_O

${I_R} = {I_T} - {I_C}$

$= {1 \over 2}{m_T}{(2R)^2} - {3 \over 2}{m_C}{R^2}$

$= {1 \over 2}(4\pi {R^2}\sigma )(4{R^2}) - {3 \over 2}(\pi {R^2}\sigma )({R^2})$

$= 6.5\pi {R^4}\sigma$

$\therefore$ ${{{I_P}} \over {{I_O}}} = {{18.5\pi {R^4}\sigma } \over {6.5\pi {R^4}\sigma }} = 2.846$

Therefore, the nearest integer is 3.