A circular disc `X` of radius `R` is made from an iron of thickness `t`, and another disc `Y` of radius `4R` is made from an iron plate of thickness `t//4`. Then the relation between the moment of mertia `I_(x)` and `I_(Y)` is :

Options

A. `I_(y)=32 I_(x)`
B. `I_(y)=16 I_(x)`
C. `I_(y)=I_(x)`
D. `I_(y)=64I_(x)`

Answer

D

Explanation

Mass of disc (X), `mx=piR^(2)t rho`
Where `rho` =density of material of disc
`:. I_(x)=1/2 m_(x)R^(2)=1/2 R^(2)t rho R^(2)`
`I_(x)=1/2 pi rho R^(4)......(i)`
Again mass of disc (Y)
`m_(y)=pi=(4R)^(2)t/4 rho=4piR^(2) t rho`
and `I_(y)=1/2 m_(y)(4R^(2))=1/2 4piR^(2)t rho.16 R^(2)`
`rArr I_(y)=32pi t rho R^(4)......(ii)`
`:. (I_(y))/(I_(X))=(32pi t rho R^(4))/(1/2 pi rho t R^(4))`
`rArr =64`
`:. :. I_(Y)=64 I_(X)`