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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • 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 ......... \(cm\).

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Answer & Solution

Correct Answer

(B) \(16\)

Step-by-step Solution

Detailed explanation

\(r = 10 cm\)                   mass per unit 

\(R = 20 cm\)                 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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