JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 22)
A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be :
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1 : 3
1 : 4
1 : 5
1 : 2
Explanation
As we know,
moment of inertia about z axis
$${{\rm I}_z} = {{m{R^2}} \over 2}$$
and moment of inertia about z'
$${\rm I}_z^1 = {3 \over 2}m{R^2}$$
$$\therefore\,\,\,\,$$ $${{{{\rm I}_z}} \over {{\rm I}{'_z}}}$$ = $${{{{m{R^2}} \over 2}} \over {{3 \over 2}m{R^2}}}$$ = $${1 \over 3}$$
moment of inertia about z axis
$${{\rm I}_z} = {{m{R^2}} \over 2}$$
and moment of inertia about z'
$${\rm I}_z^1 = {3 \over 2}m{R^2}$$
$$\therefore\,\,\,\,$$ $${{{{\rm I}_z}} \over {{\rm I}{'_z}}}$$ = $${{{{m{R^2}} \over 2}} \over {{3 \over 2}m{R^2}}}$$ = $${1 \over 3}$$
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