A circular disc X of radius 'R' is made from an iron plate of thicknes

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A circular disc X of radius 'R' is made from an iron plate of thickness 't', and another disc Y of radius '4R' is made from an iron plate of thickness '

\frac{t}{4}

'. Then the relation between the moment of inertia

I_{x}

and

I_{y}

is

Options:

A . Iy=64Ix

B . Iy=32Ix

C . Iy=16Ix

D . Iy=Ix

Answer: Option A
:
A
Moment of Inertia of disc

I = \frac{1}{2} M R^{2} = \frac{1}{2} (π R^{2} t ρ) R^{2} = \frac{1}{2} π t ρ R^{4}

[As

M = V \times ρ = π R^{2} t ρ

where t = thickness,

ρ

= density]

∴ \frac{I_{y}}{I_{x}} = \frac{t_{y}}{t_{x}} {(\frac{R_{y}}{R_{x}})}^{4}

[ If

ρ

= constant]

\Rightarrow \frac{I_{y}}{I_{x}} = \frac{1}{4} (4)^{4} = 64

[G i v e n R_{y} = 4 R_{x}, t_{y} = \frac{t_{x}}{4}]

\Rightarrow I_{y} = 64 I_{x}

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