- JEE Mains 2019
- Hard
- Numerical
Let the moment of inertia of a hollow cylinder of length \(30\, cm\) (inner radius \(10\, cm\) and outer radius \(20\, cm\)), about its axis be \(I\). The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also \(I\), is ......... \(cm\).
Answer & Solution
(B) \(16\)
Step-by-step Solution
Detailed explanation
\(R = 20 cm\) area
\(Mass = m\) \(\sigma = \frac{m}{{\pi \left( {{R^2} - {r^2}} \right)}}\)
Consider an element of radius \(x\) and thickness \(dx\)
Mass of element \(dm =\) \(\sigma \,2\pi \times x\left( {dx} \right)\)
Moment of inertia of element, \(dI = (dm)\)\({x^2}\)
\( \Rightarrow \,\,I = \sigma 2\pi \int_r^R {{X^3}dx} \)
\( = \sigma \frac{{2\pi }}{4}\left( {{R^4} - {r^4}} \right)\)
\( = \frac{m}{{\pi \left( {{R^2} - {r^2}} \right)}}\frac{\pi }{2}\left( {{R^4} - {r^4}} \right)\)
\(\,I = \frac{m}{2}\left( {{R^2} + {r^2}} \right)\,\,\,\,\,\,\,\,\,...\left( i \right)\)
Moment of inertia of thin cylinder of same mass,
\(I = mr_0^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\)
\( \Rightarrow \,\,mr_0^2 = \frac{m}{2}\left( {{R^2} + {r^2}} \right)\)
\(r_0^2 = 250\,\,\,\,\,\,;\,\,\,\,\,\,{r_0}\, \approx \,16\,cm.\)
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