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

Options

A. `I_(Y)=32I_(X)`
B. `I_(Y)=16I_(X)`
C. `I_(Y)=I_(X)`
D. `I_(Y)=64I_(X)`

Answer

D

Explanation

`I=(1)/(2)mR^(2)MproptpropR^(2)`
For disc `X,I_(X)=(1)/(2)(m)(R)^(2)=(1)/(2)(pir^(2)t).(R)^(2)`
for disc `Y,I_(Y)=(1)/(2)[pi(4R)^(2).t//4][4R]^(2)`
`implies(I_(X))/(I_(Y))=(1)/((4)^(3))impliesI_(Y)=64I_(X)`