A thin wire of length / having a linear density p is bent in to a circular loop with c as its centre as shown in the figure the moment of inertial of the loop about the line AB is

Options

A. `(5)/(16) .(pl^(3))/(pi^(3))`
B. `(1)/(16) (pl^(3))/(pi^(2))`
C. `(1)/(8) (pl^(3))/(pi^(3))`
D. `(3)/(8) (pl^(3))/(pi^(2))`

Answer

D

Explanation

Given length of a thin wire = l
Linear density fo the wire = p
Let the mass of the wire = M ,
Now according to the question

Here circular loop of the wire with centre C R is radius of circular loo of wire
`:.` Moment of inertia axis passing through the centre and perpendicular to the plane AB .
Now the total moment of inerial for circular loop of wire .If R is the distance between the axis `I_(CM)` and `I_(AB)` are the respective moments of inertia about these axies .
Then `I = l_(CM) + I_(AB)` From Eqs . (i) and (ii) we get