A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1:3, then the force constant of smaller piece is __ /m.

We need to find the force constant of the smaller piece when a spring of force constant 15 N/m is cut in the ratio 1:3.

Since the force constant k of a spring of length L is inversely proportional to its length, we have $$k \times L = k_1 \times L_1 = k_2 \times L_2 = \text{constant}$$

Given that k = 15 N/m and the spring is cut in the ratio 1:3, if the total length is L then the smaller piece has length $$L_1 = \frac{L}{4}$$ and the larger piece has length $$L_2 = \frac{3L}{4}$$.

Substituting into the relation gives $$k_1 = \frac{k \times L}{L_1} = \frac{15 \times L}{\frac{L}{4}} = 15 \times 4 = 60 \text{ N/m}$$

Therefore, the force constant of the smaller piece is 60 N/m (Option 3).