A spring of force constant k is cut into length of ratio 1 : 2 : 3. They are connected in series and the new force constantis k'. Then they are connected in parallel and force constant is k''. That k' : k'' = …………..

A spring of force constant k is cut into length of ratio 1 : 2 : 3. Th

1.	A spring of force constant k is cut into length of ratio 1 : 2 : 3. They are connected in series and the new force constantis k'. Then they are connected in parallel and force constant is k''. That k' : k'' = …………..
Answer» Solution :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)`.