Three point masses are at the corners of an equilateral traingle of side `r`. Their separations do not change when the system rotates about the centre of the triangle. For this, the time period of rotation must be proportional to

Options

A. `sqrt(L)`
B. `L^(3//2)`
C. L
D. `L^(-2)`

Answer

B

Explanation

The situation can be shown below.

where O is the centriod of equivalent triangle.
The length AO can be found by
`AO = sqrt(OD^(2) + AD^(2))`
`= sqrt(((1)/(3)BD^(2)) +AD^(2)) = sqrt((1)/(9) xx (sqrt(3)/(2)L)^(2) +(L^(2))/(4))`
`=sqrt((1)/(12)L^(2)+(L^(2))/(4))=(L)/(sqrt(3))`
The moment of inertia about O is given by
`I = MR^(2)`
` = m xx (L)/(sqrt(3))^(2) = (mL^(2))/(3)" ".......(ii) `
According to law of conervation of angular momentum,
`I omega` = constant
`I xx (2pi)/(T)` = constant
`rAr " " T prop l`
From Eq.(i) , we get
`T prop L^(2)`