Question

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

• A. I_Y = 64I_X
• B. I_Y = 32I_X
• C. I_Y = 16I_X
• D. I_Y = I_X

Answer 1

Answer: (A)

I_Y = 64I_X

Answer 2

Moment of Inertia of disc I = $\frac { 1 } { 2 } M R ^ { 2 }$ $= \frac { 1 } { 2 } \left( \pi R ^ { 2 } t \rho \right) R ^ { 2 }$ $= \frac { 1 } { 2 } \pi t \rho R ^ { 4 }$

[As $M = V \times \rho$= where $t =$ thickness, ρ = density]

∴ $\frac { I _ { y } } { I _ { x } } = \frac { t _ { y } } { t _ { x } } \left( \frac { R _ { y } } { R _ { x } } \right) ^ { 4 }$

[If $\rho$ = constant]

⇒ $\frac { I _ { y } } { I _ { x } } = \frac { 1 } { 4 } ( 4 ) ^ { 4 } = 64$ [Given$R _ { y } = 4 R _ { x }$, $t _ { y } = \frac { t _ { x } } { 4 }$]

⇒ $I _ { y } = 64 I _ { x }$