A spring of force constant k is cut into length 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''. That k' : k'' = …………..

Answer

B

Explanation

Suppose spring of length l is cut in the ratio 1 : 2 : 3.
The length of spring segments `= (l)/(6),(2l)/(6),(3l)/(6)` respectively.
Now, `k propto (l)/(l)`
`(k_1)/(k)= (l)/(l_1) implies k_(1)= (kl)/(l_1)= (kl)/((l)/(6))= 6k`
and `(k_2)/(k)= (l)/(l_2) implies k_(2)= (kl)/(l_2)= (kl)/((l)/(3))= 3k`
and `(k_3)/(k)= (l)/(l_3) implies k_(3)= (kl)/(l_3)= (kl)/((l)/(2))= 2k`
For series connection `(1)/(k.)= (1)/(6k) + (1)/(3k)+ (1)/(2k) = (1)/(k)`
`therefore k.= k" ""........."(1)`
and for parallel connection `k..= 6k+3k+2k`
`therefore k..= 11k " ""........"(2)`
`therefore (k.)/(lk..)= (1)/(11)`.