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QJEEM20Ph8bab: JEE Main 2020 Physics - Class 11 JEEM20Ph8bab Solution: Three solid spheres each of mass m and diameter d are stuck…

JEE Main 2020 (Online) 9th January Morning Slot Physics - Rotational Motion Question 138 English 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 :

Multiple Choice Options

  • A 1323{{13} \over {23}}
  • B 2313{{23} \over {13}}
  • C 1513{{15} \over {13}}
  • D 1315{{13} \over {15}}

Step-by-Step Verified Solution

AB = d

∴\therefore OA = d3{{d \over {\sqrt 3 }}}

From parallel axis theorem

I0 = 3×[25M(d2)2+M(d3)2]3 \times \left[ {{2 \over 5}M{{\left( {{d \over 2}} \right)}^2} + M{{\left( {{d \over {\sqrt 3 }}} \right)}^2}} \right] = 1310Md2{{13} \over {10}}M{d^2}

IA = I0 + 3M (d3)2{{{\left( {{d \over {\sqrt 3 }}} \right)}^2}}

= 1310Md2{{13} \over {10}}M{d^2} + Md2

= 2310Md2{{23} \over {10}}M{d^2}

∴\therefore I0IA{{{I_0}} \over {{I_A}}} = 1323{{13} \over {23}}

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