the density of a rod AB increases linearly from A to B. mid point of the rod is O and its centre of mass is at C. four axes pass through A,B,O and C. the moment of in ertia of the rod about these axes are IA, IB, IO and IC respectively. Then
A IA = IB
B IA < IB
C IO > IC
D IO < IC
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The correct option is (c) Io > Ic.
Given-
- Density increases linearly from A to B.
- Moment of inertia about A, B, O and C axis are Ia, Ib, Io and Ic
Axis are passes from all the points as shown in attached figure.
AO is less denser side because density increases linearly so center of mass lies after O as OB will be more denser side than AO.
From parallel axis theorem
Ia = Ic + m(AC)² where AC is the distance between A and C.
- Similarly we can write the moment of inertia for Io and Ib in terms of C.
- Ic will be least because we know that moment of inertia that passes from the center is least.
Moment of inertia is proportional to mass and square of the distance.
- Ia will be more than Ib because density increases linearly means mass near to the A point will be in more distance as compared to mass present in B point.
Hence correct option is (c) Io > Ic
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Answer : IA>IB
Explanation
Ic is least and C lies between O and B as density is increasing from A to B
as more mass is concentrated towards B
so the answer is IA > IB .
HOPE IT HELPS ..
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