When the temperature of a body increases from `T to T+ DeltaT`, its moment of inertia increases from `I to I + DeltaI`. If `alpha` is the coefficient of linear expansion of the material of the body, the `(Delta I)/(I)`is (neglect higher order of `alpha`)

When the temperature of a rod increases from t to t+Delta t , its moment of inertia increases from I to I+Delta I . If alpha is coefficient of linear expansion, the value of Delta I//I is

When the temperature of a rod increases from t to r+Delta t , its moment of inertia increases from I to I+Delta I . If alpha is the value of Delta I//I is

When the temperature of a body increases from t to t+Deltat , the moment of inertia of it increases from I to I+DeltaI . If the coefficient of linear expansion of the body is alpha, then show that (DeltaI)/I=2alphaDeltat .

When the temperature of a rod increases from 1 to t. Delta t,the moment of inertia increases from 1 to 1+Delta.The coefficient of linear expansion of the rod is alpha.The ratio Delta I/I is :

when the temperature of a body increses from t to t +Deltat ,its moment of inertia increases form I to I + DeltaI . The coefficient of linear expanison of the body is prop . The ratio (DeltaI)/I is equal to

When temperature of a body made of brass increases its moment of inertia.

The moment of ineratia of a uniform thin rod about its perpendicular bisector is I . If the temperature of the rod is increased by Deltat , the moment of inertia about perpendicular bisector increases by (coefficient of linear expansion of material of the rod is alpha ).

If the temperature of a unifrom rod is slightly increased by Deltat , its moment of inertia I about a line parallel to itself will increase by