A block of weight W produces an extension of 9cm when it is hung by an elastic spring of length 60 cm and is in equilibrium. The spring is cut into two parts, one of length 40 cm and the other of length 20cm. The same load W hangs in equilibrium supported by both parts as shown in fig(b). The extension in cm now is
 

If K be the spring constant, then
$\mathrm{W}=\mathrm{kx}\text{ or }\mathrm{W}=9\mathrm{k}$…………(1)
We know that spring constant is inversely proportional to length. Hence spring constant for spring of length no cm is (3/2)K and that for 20 cm length is 3 k. The combined spring constant for second case is given by
${\mathrm{k}}^{\mathrm{\prime }}=\frac{3}{2}\mathrm{k}+3\mathrm{k}=\frac{9\mathrm{k}}{2}$
In this position,
$\mathrm{W}=\left(\frac{9\mathrm{k}}{2}\right)\times {\mathrm{x}}^{\mathrm{\prime }}$…………….(2)
From eqs. (1) and (2), we get
$9\mathrm{k}=\left(\frac{9\mathrm{k}}{2}\right)\times {\mathrm{x}}^{\mathrm{\prime }}$…………(2)
or  x' = 2cm