Rotational Motion

JEE Main level, numerical answer, medium

21 of 280

Question 21 of 280

A,B\mathrm{A}, \mathrm{B} and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.

The moment of inertia of the given system about
PQP Q is X15I\frac{X}{15} I, where II is the moment of inertia of the disc about its diameter. The value of x is _______.

Show answer and solution

Answer: 199


All bodies have same mass and same radius.

A→\mathrm{A} \rightarrow Disc
B→\mathrm{B} \rightarrow Solid sphere
C→\mathrm{C} \rightarrow Spherical shell
and,
I=MR24\mathrm{I}=\frac{\mathrm{MR}^2}{4}
IPQ=MR24+(25MR2+MR2)+(23MR2+MR2)\mathrm{I}_{\mathrm{PQ}}=\frac{\mathrm{MR}^2}{4}+\left(\frac{2}{5} \mathrm{MR}^2+\mathrm{MR}^2\right)+\left(\frac{2}{3} \mathrm{MR}^2+\mathrm{MR}^2\right)
IPQ=15MR2+24MR2+60MR2+40MR2+60MR260\mathrm{I}_{\mathrm{PQ}}=\frac{15 \mathrm{MR}^2+24 \mathrm{MR}^2+60 \mathrm{MR}^2+40 \mathrm{MR}^2+60 \mathrm{MR}^2}{60}
IPQ=19960MR2=19915(MR24)\mathrm{I}_{\mathrm{PQ}}=\frac{199}{60} \mathrm{MR}^2=\frac{199}{15}\left(\frac{\mathrm{MR}^2}{4}\right)
=19915I=\frac{199}{15} \mathrm{I}

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