Q 11-13-019NEETJEE MainEasy
Two springs of force constants $k_1 = 200$ N/m and $k_2 = 300$ N/m are joined in series. The force constant of the combination is
Answer: (A) $120$ N/m
Concept. Springs in series: $\dfrac1k = \dfrac{1}{k_1} + \dfrac{1}{k_2}$.
Step 1. Substitute. $\dfrac1k = \dfrac{1}{200} + \dfrac{1}{300} = \dfrac{5}{600}$.
Step 2. Result. $k = 120$ N/m.
Answer. $120$ N/m (Option A).
Solution by Sreeraj P, M.Sc Physics