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

Answer

B

Explanation

Let us assume, the length of spring be l.
When we cut the spring into ratio of length `1:2:3`, we get three springs of lengths `(l)/(6),(2l)/(6) and (3l)/(6)` with force cosntant,
`therefore k_(1)=(kl)/(l_(1))=(kl)/(l//6)=6k,k_(2)=(kl)/(l_(2))=(kl)/(2l//6)=3k`
`k_(3)=(kl)/(l_(3))=(kl)/(3l//6)=2k`
When connected in series,
`(1)/(k.)=(1)/(6k)+(1)/(3k)+(1)/(2k)=(1+2+3)/(6k)=(1)/(k)`
`therefore k.=k`
When connected in parallel, `k..=6k+3k+2k=11k`
`(k.)/(k..)=(k)/(11k)=(1)/(11)`