JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 7)
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Three solid spheres each of mass m and
diameter d are stuck together such that the lines
connecting the centres form an equilateral
triangle of side of length d. The ratio I0/IA of
moment of inertia I0 of the system about an axis
passing the centroid and about center of any
of the spheres IA and perpendicular to the plane
of the triangle is :$${{13} \over {23}}$$
$${{23} \over {13}}$$
$${{15} \over {13}}$$
$${{13} \over {15}}$$
Explanation
AB = d
$$ \therefore $$ OA = $${{d \over {\sqrt 3 }}}$$
From parallel axis theorem
I0 = $$3 \times \left[ {{2 \over 5}M{{\left( {{d \over 2}} \right)}^2} + M{{\left( {{d \over {\sqrt 3 }}} \right)}^2}} \right]$$ = $${{13} \over {10}}M{d^2}$$
IA = I0 + 3M$${{{\left( {{d \over {\sqrt 3 }}} \right)}^2}}$$
= $${{13} \over {10}}M{d^2}$$ + Md2
= $${{23} \over {10}}M{d^2}$$
$$ \therefore $$ $${{{I_0}} \over {{I_A}}}$$ = $${{13} \over {23}}$$
$$ \therefore $$ OA = $${{d \over {\sqrt 3 }}}$$
From parallel axis theorem
I0 = $$3 \times \left[ {{2 \over 5}M{{\left( {{d \over 2}} \right)}^2} + M{{\left( {{d \over {\sqrt 3 }}} \right)}^2}} \right]$$ = $${{13} \over {10}}M{d^2}$$
IA = I0 + 3M$${{{\left( {{d \over {\sqrt 3 }}} \right)}^2}}$$
= $${{13} \over {10}}M{d^2}$$ + Md2
= $${{23} \over {10}}M{d^2}$$
$$ \therefore $$ $${{{I_0}} \over {{I_A}}}$$ = $${{13} \over {23}}$$
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