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

Answer

2

Explanation

`mg=(K)(9)orK=mg//9`
The spring constant is inversely proportional to length. So the spring constant for the spring of 40 cm length is `(3)/(2)`K.
The spring constant for the spring of 20 cm length is 3 K. The spring constant for the combination is
`(3)/(2)K+3K=(9)/(2)K`, Now `mg=(9)/(2)KX_(0)or9K=(9)/(2)KX_(0)X_(0)=2m`