A spring whose unstretched length is $$l$$ has a force constant k. The spring is cut into two pieces of unstretched lengths $$l_1$$ and $$l_2$$ where, $$l_1 = nl_2$$ and n is an integer. The ratio $$k_1/k_2$$ of the corresponding force constants, k$$_1$$ and k$$_2$$ will be:

We begin with the basic property of a uniform spring: its force constant (spring constant) $$k$$ is inversely proportional to its natural or unstretched length $$l$$. In other words, for any piece cut from the same spring, the product $$k \, l$$ remains constant. We can write this as

$$k \, l = \text{constant} \qquad \text{(for a given uniform spring)}$$

Let this constant be denoted by $$C$$, so that for the original full spring we have

$$k \, l = C \quad -(1)$$

Now the spring is cut into two parts. Their natural lengths are stated to be $$l_1$$ and $$l_2$$, and the problem tells us that

$$l_1 = n \, l_2 \qquad \text{where } n \text{ is an integer.} \quad -(2)$$

Because both pieces come from the same original spring, each piece must satisfy the same relation (1). Hence for the first piece we can write

$$k_1 \, l_1 = C \quad -(3)$$

and for the second piece we can similarly write

$$k_2 \, l_2 = C. \quad -(4)$$

From equation (3) we isolate $$k_1$$:

$$k_1 = \dfrac{C}{l_1}. \quad -(5)$$

From equation (4) we isolate $$k_2$$:

$$k_2 = \dfrac{C}{l_2}. \quad -(6)$$

We are asked to find the ratio $$\dfrac{k_1}{k_2}$$. Dividing (5) by (6) we get

$$\dfrac{k_1}{k_2} = \dfrac{C/l_1}{C/l_2} = \dfrac{l_2}{l_1}. \quad -(7)$$

Now we substitute the relation between the lengths from equation (2). Because $$l_1 = n l_2$$, we can solve for $$l_2$$:

$$l_2 = \dfrac{l_1}{n}. \quad -(8)$$

Putting (8) into (7) gives

$$\dfrac{k_1}{k_2} = \dfrac{l_2}{l_1} = \dfrac{l_1/n}{l_1} = \dfrac{1}{n}. \quad -(9)$$

Thus the ratio of the spring constants of the two pieces is

$$\dfrac{k_1}{k_2} = \dfrac{1}{n}.$$

Comparing with the options provided, this matches Option D.

Hence, the correct answer is Option D.