Correct Answer - B
Let x be the maximum extension of the spring. When it is loaded with a mass M and released. There is a decrease in the gravitational P.E. and it is = Mgx
But this increases P.E. of the spring
`therefore P.E. =(1)/(2)Kx^(2) " " therefore (1)/(2)Kx^(2)=Mgx " " therefore x = (2Mg)/(K)`