A spring of force constant `k` is cut into lengths of ratio `1 : 2 : 3`. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k'. Then k' : k" is :

Options

A. `1:6`
B. `1:9`
C. `1:11`
D. `1:14`

Answer

C

Explanation

Let `k_(1), k_(2)` and `k_(3)` be the force constant of three lengths of a spring into which a spring is cut. As spring constant `prop (1)/(l e n g t h)`
So, `k_(1):k_(2):k_(3)=(1)/(l_(1)):(1)/(l_(2)):(1)/(l_(3))=(1)/(1):(1)/(2):(1)/(3)=(6:3:2)/(6)`
`,. k_(1)=6K:k_(2)=3K` and `k_(3)=2K`
In series combination,
`(1)/(k)=(1)/(6K)+(1)/(3K)+(1)/(2K)=(6)/(6K)` or `k^(')=K`
In parallel combination,
`k^('')=6K+3K+2K=11K`
`:. (k^('))/(k^(''))=(K)/(11K)=(1)/(11)`