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
$$\frac{1}{k} = \frac{1}{200} + \frac{1}{300} = \frac{5}{600} \;\Rightarrow\; k = 120\ \text{N/m}$$
Solution by Sreeraj P, M.Sc Physics