A massless platform is kept on a light elastic spring as shown in the figure. When a particle of mass 0.1 kg is dropped on the pan from a height of 0.24m, the particle strikes the pan and the spring is compressed by 0.01m. From what height should the particle be dropped to cause a compression of 0.04m?

Let the particle be dropped from a height h and the spring be compressed by y. According to the conservation of mechanical energy, loss in PE of the particle = gain in elastic potential energy of the spring 
         $mg (h+y) = \frac{1}{2} k{y}^2$, here m is mass of particle, g is acceleration due to gravity, h is height, y is extent of compression, k is force constant
Now, as the particle and spring are same in the  second case also,
   $\frac{mg(h_1+y_1)}{mg(h_2+y_2)}=\frac{{\displaystyle \frac{1}{2}}k{y_1}^2}{\frac{1}{2}k{y_2}^2}$ or 

  $\left(\frac{0.24+0.01}{h_2+0.04}\right)={\left(\frac{0.01}{0.04}\right)}^2$
Solving we get $h_2 = 3.96 m$