1.

The moment of inertia of a uniform rod about a perpendicular axis passing through one end is `I_(1)`. The same rod is bent into a ring and its moment of inertia about a diameter is `I_(2)`. Then `I_(1)//I_(2)` isA. `(pi)/(3)`B. `(8pi^(2))/(3)`C. `(5pi)/(3)`D. `(8pi^(2))/(5)`

Answer» Correct Answer - B
`I=(mL^(2))/(3), I_(1)=(mR^(2))/(2)`
`(I)/(I_(1))=(mL^(2))/(3)xx(2)/(MR^(2))`
`= (2)/(3)xx(L^(2))/(R^(2))`
`=(2)/(3)xx((2pi R^(2)))/(R^(2)) because L = 2 pi R`
`= (2)/(3)xx (4pi^(2)R^(2))/(R^(2))`
`(I)/(I_(1))=(8pi^(2))/(3)`.


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