Two rods one is hollow and other is solid of same length and made of same material having same modulus of rigidity. Prove that hollow rod is stronger than solid rod.

Explanation

Consider a solid rod of radius (r) and length.(l). Let a twisting couple is applied on its free end so that lower face is twisted then torque per unit twist is given by
`tau=(pi eta r^4)/(2l)" "`...(i)
also consider a hollow rod of inner radius `(r_(1))` and outer radius `(r_2)` then twisting torque per unit twist is given by
`tau.=(pi eta)/(2l)(r_2^4-r_(1)^4)" "`.....(ii)
Dividing (ii) by (i), we have
`(tau.)/(tau)=(r_2^4-r_(1)^4)/(r^4)=((r_2^2-r_(1)^2)(r_2^2+r_(1)^2))/((r^2)^2)" "`.......(iii)
Since, mass of both the rods is same,
then `pi r^2 l rho=pi(r_2^2-r_(1)^2)l rho` where `rho` is density of material of rods.
or `r^2=r_2^2-r_(1)^2" "`........(iv)
then using (iv) in (iii), we get
`(tau.)/(tau)=((r_2^2+r_(1)^2))/(r^2) gt 1`
or `tau. gt tau`
Hence, hollow rod is stronger than solid rod of same length and made of same material having same modulus of rigidity.