NEET2017PhysicsLaws of MotionActual
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 ′ . If now they are connected in parallel and force constant is k ′ ′ , then   k '   :   k ' ' is
Options
- A1   :   6
- B1   :   9
- C1   :   11
- D1   :   14
Correct answer
C. 1   :   11
Step-by-step solution
Spring constant ∝ 1 l e n g t h k ∝ 1 l , Spring constant for the length ratios are k 1 = 6 k k 2 = 3 k k 3 = 2 k In series connection of springs, 1 k ′ = 1 6 k + 1 3 k + 1 2 k 1 k ′ = 6 6 k k ′ = k . In parallel connection of springs, k ' ′ = 6 k + 3 k + 2 k k ′ ′ = 11 k k ′ k ′ ′ = 1 11 i.e., k ′   : k ′ ′ = 1   : 11