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Rotational Motion question

2019 · 12 Jan · Shift 2 · Q58

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Rotational Motion
2019 · 12 Jan · Shift 2 · Q58

Rotational Motion question

2019 · 12 Jan · Shift 2 · Q58

JEE MainPhysicsRotational MotionMCQ+4 / −1

The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly ?

A
B
C
D

View written solutionFree

Correct answer: C

Moment of inertia of a solid sphere about a diameter

For a solid sphere of mass $M$ and radius $R$ , the moment of inertia about any diameter is

$I_0 = \frac{2}{5}MR^2.$

Axis parallel to the diameter and at distance $x$

If the new axis is parallel to that diameter and shifted by a perpendicular distance $x$ , then by the parallel axis theorem,

$I(x) = I_0 + Mx^2.$

So,

$I(x) = \frac{2}{5}MR^2 + Mx^2.$

Nature of the graph

This is of the form

$I(x) = \text{constant} + Mx^2,$

which is:

a parabola opening upward,
symmetric about $x=0$ ,
having minimum value at $x=0$ ,
and

$I(0)=\frac{2}{5}MR^2 \neq 0.$

Thus the graph should be an upward-opening parabola with a positive intercept on the $I$ -axis.

Correct option

Hence the correct graph is the one representing

$I(x)=\frac{2}{5}MR^2+Mx^2,$

i.e. Option C.

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