A rod of uniform cross-section of mass `M` and length `L` is hinged about an end to swing freely in a vertical plane. However, its density is non uniform and varies linearly from hinged end to the free end doubling its value. The moment of inertia of the rod, about the rotation axis passing through the hinge point

Options

A. `2(ML^(2))/(9)`
B. `(3ML^(2))/(16)`
C. `(7ML^(2))/(18)`
D. none of these

Answer

C

Explanation

`S=` area of cross-section
`rho_(x)=rho_(0)+((2rho_(0)-rho_(0))/(L))x=rho_(0)+(rho_(0))/(L)x`
Now, `M=int_(0)^(L)dM=int_(0)^(L)(Sdx)(rho_(0)+(rho_(0))/(L)x)`
`=rho_(0)S((3L)/(2))`
`thereforerho_(0)S=(2M)/(3L)` ..(i)
Now, `I=int_(0)^(L)dI=int_(0)^(L)(dM)x^(2)`
`=int_(0)^(L)(Sdx)(rho_(0)+(rho_(0))/(L)x)(x^(2))`
After substituting velue of `rho_(0)S` from eq (i) then we find
`I=(7)/(18)ML^(2)`