A thin wire of length `L` and uniform linear mass density `rho` is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis `XX'` is `A` cubical block of side `L` rests on a rough horizontal surface with coefficient of friction sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

Options

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

Answer

D

Explanation

The moment of inertia of the loop about `XX'` axis is
`I_(XX)=(mR^(2))/2+mR^(2)+3/2mR^(2)`
where `m=` mass of the loop and `R=`radius of the loop
Here `m-L rho` and `R =L/(2pi)`, therefore
`I_(XX')=3/2(Lrhop)(L/(2pi))^(2)=(3L^(3)rho)/(8pi^(2))`