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 shown in figure. The extension in `cm` is

Options

A. 9
B. 6
C. 5
D. 2

Answer

D

Explanation

(d) Force constant ,`kprop(1)/(l)`
Let k be the force constant of original spring.
Then, `k_(40)=(60)/(40)k=(3)/(2)k" and "k_(20)=(60)/(20)k=3k`
`k_(net)=k_(40)+k_(20)=(9k)/(2)`
Now, `Deltal=(F)/(k) prop (1)/(k)`
`therefore Deltal'=((k)/(9k//2))(9 cm)=2 cm`