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

Options

A. 2
B. 3
C. 6
D. 9

Answer

A

Explanation

Let `k` be the spring constant of original spring.
Then `W=kxx9` or `k=(W)/(9)`
As spring constant is inversely proportional to length, so ration of spring constant of length `20 cm` and `40cm` is `(k_(1))/(k_(2))=(2)/(1)` or `k_(1)=2k_(2)`
When these two parts of spring act as one spring of spring constant `k`, then
`(k_(1)k_(2))/(k_(1)+k_(2))=k` or `(2k_(2)xxk_(2))/(2k_(2)+k_(2))=k` or `k_(2)=(3)/(2)k`
and `k_(1)=2xx(3)/(2)k=3k`
When shorter spring are connected as shown in fig, the total spring constant is
`K=3k+(3k)/(2)=(9k)/(2)`
In equilibrium position, `W=(9)/(2)kxxx^(')`
or `9k=(9)/(2)kx^(')` or `x^(')=2cm`