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A circular disc X of radius R is made fr...

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 `(t)/(4)`. Then the relation between the moment of inerita `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)`

Text Solution

Verified by Experts

The correct Answer is:
A
Moment of Inertia of disc `l=(1)/(2)MR^(2)=(1)/(2)(piR^(2)trho)R^(2)=(1)/(2)pitrhoR^(4)` [As `M=Vxxrho=piR^()trho` where t = thickness, `rho` = density]
`therefore (I_(y))/(I_(x))=(t_(y))/(t_(x))((R_(y))/(R_(x)))^(4)` [If `rho` = constant]
`implies (I_(y))/(I_(x))=(1)/(4)(4)^(4)=64` [Given `R_(y)=4R_(x),t_(y)=(t_(x))/(4)`]
`implies I_(y)=64I_(x)`
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