JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 18)
Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as;
I1 = M.I. of thin circular ring about its diameter,
I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I3 = M.I. of solid cylinder about its axis and
I4 = M.I. of solid sphere about its diameter.
Then :
I1 = M.I. of thin circular ring about its diameter,
I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I3 = M.I. of solid cylinder about its axis and
I4 = M.I. of solid sphere about its diameter.
Then :
I1 = I2 = I3 > I4
I1 + I3 < I2 + I4
I1 = I2 = I3 < I4
I1 + I2 = I3 + $${5 \over 2}$$ I4
Explanation
Let M and R be the mass and radius of four bodies. Then, as per
question, their moment of inertia are
I1 = $${{M{R^2}} \over 2}$$,
I2 = $${{M{R^2}} \over 2}$$,
I3 = $${{M{R^2}} \over 2}$$,
I4 = $${2 \over 5}M{R^2}$$
$$ \therefore $$ I1 = I2 = I3 > I4
I1 = $${{M{R^2}} \over 2}$$,
I2 = $${{M{R^2}} \over 2}$$,
I3 = $${{M{R^2}} \over 2}$$,
I4 = $${2 \over 5}M{R^2}$$
$$ \therefore $$ I1 = I2 = I3 > I4
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