Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as -

Question:

Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as -

$\mathrm{I}_{1}=$ M.I. of thin circular ring about its diameter,

$\mathrm{I}_{2}=$ M.I. of circular disc about an axis perpendicular to disc and going through the centre,

$\mathrm{I}_{3}=$ M.I. of solid cylinder about its axis and

$\mathrm{I}_{4}=$ M.I. of solid sphere about its diameter.

Then :-

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

  2. $\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{3}+\frac{5}{2} \mathrm{~L}_{4}$

  3. $\mathrm{I}_{1}+\mathrm{I}_{3}<\mathrm{I}_{2}+\mathrm{I}_{4}$

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


Correct Option: , 4

Solution:

(4)

Given $\Rightarrow I_{1}=M . I$. of thin circular ring about its diameter $\mathrm{I}_{2}=$ M.I. circular disc about an axis perpendicular to disc and going through the centre. $I_{3}=M . I$. of solid cylinder about its axis

$\mathrm{I}_{4}=$ M. I, of solid sphere about its diameter we know that,

$I_{1}=\frac{M R^{2}}{2}, I_{2}=\frac{M R^{2}}{2}, I_{3}=\frac{M R^{2}}{2}$

$I_{4}=\frac{2}{5} M R^{2}$

So, $I_{1}=I_{2}=I_{3}>I_{4}$

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