Find the `M.I.`

1.	Find the `M.I.`
Answer» `Mass//area = (m)/(pi(r_(1)^(2)-r_(2)^(2)))` Taking a ring of radius `r` and thickness `dr` `dA=2pidr` `dm=(m)/(pi(r_(1)^(2)-r_(2)^(2)))xx2pirdr=(2mrdr)/((r_(1)^(2)-r_(2)^(2)))` `I= intr^(2)dm=(2m)/((r_(1)^(2)-r_(2)^(2))) int_(r_(2))^(r_(1))r^(3) dr` ` =(2m)/((r_(1)^(2)-r_(2)^(2)))\|(r_(4))/(4)\|_(r_(2))^(r_(1))` `= (1)/(2)(m(r_(1)^(4)-r_(2)^(4)))/((r_(1)^(2)-r_(2)^(2)))=(1)/(2)(m(r_(1)^(2)+r_(2)^(2))(r_(1)^(2)-r_(2)^(2)))/((r_(1)^(2)-r_(2)^(2)))` `=(1)/(2)m(r_(1)^(2)+r_(2)^(2))`

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