d) from energy conservation,
`M/2gl=1/2lw^(2)`………….(i)
Moment of inertia of the rod,
`l=1/3 Ml^(2)`………….(ii)
From eqs (i) and (ii), we get
`M/2gl = 1/2(Ml^(2))/3 omega^(2) rArr omega=sqrt(3g)/l`
Given, l =1m, `omega=sqrt(g)`
Also, `v=romega`= `1 xx sqrt(3xx9.8) = 5.4 m//s`