JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 7)
A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of $${{7M} \over 8}$$
and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere.
Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere
about its axis. The ratio I1/I2 is given by :
65
140
185
285
Explanation
$${I_1} = {{\left( {{{7M} \over 8}} \right){{\left( {2R} \right)}^2}} \over 2} = {{7M \times 4{R^2}} \over {2 \times 8}} = {{7M{R^2}} \over 4}$$
$${I_2} = {2 \over 5}{M \over 8}{\left( {{R \over 2}} \right)^2} = {{2M} \over {5 \times 8}}{{{R^2}} \over 4} = {{M{R^2}} \over {80}}$$
$${{{I_1}} \over {{I_2}}} = {{7M{R^2} \times 80} \over {4M{R^2}}} = 140$$
$${I_2} = {2 \over 5}{M \over 8}{\left( {{R \over 2}} \right)^2} = {{2M} \over {5 \times 8}}{{{R^2}} \over 4} = {{M{R^2}} \over {80}}$$
$${{{I_1}} \over {{I_2}}} = {{7M{R^2} \times 80} \over {4M{R^2}}} = 140$$
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