A thin uniform rod of length `L = 100cm` is to be pivoted about some point passing through the rod is allowed to swing as a physical pendulum. For this situation mark the correct statement (s) [take `g = 10m//s^(2)]`

Options

A. If the rod is pioted at `29cm` (approximately) away form centre of rod then it will perform `SHM` with least time period
B. If the rod is pivoted at `15cm` (approximately) away form centre of rod then it will performs `SHM` with least time period.
C. The least time period (approximately) with which rod can oscillate is `1.5s`
D. There are two possible loaction of pivot about which rod can perform `SHM` with leats time period.

Answer

A, C, D

Explanation

`T = 2pi sqrt((I)/(mgd))`
where `I` is momet of inertia of road about `P`.

i.e. `I = (mL^(2))/(12) +md^(2)`
For minimum time period `(dT)/(dd) = 0`
`d = (l)/(sqrt(12)) = 29 cm`
Value of minimum time `T = 1.5 sec`