Answer:

Stiffness of a spring is defined as the force required to produce a unit deformation in the spring.

When a closely-coiled helical spring is cut into two halves, the following changes occur:

1. Length of each half of the spring becomes half of the original length.
2. The number of turns in each half of the spring becomes half of the original number of turns.
3. The wire diameter and the material of the spring remain the same.

Therefore, the stiffness of the resulting spring can be determined as follows:

Stiffness ∝ (Number of turns)⁴/(Wire diameter)³

Let's consider two cases:

Case 1: When the original spring is cut into two halves, and each half is used as a separate spring

In this case, the stiffness of each half of the spring will be the same as the stiffness of the original spring because the wire diameter and the material of the spring are the same.

Therefore, the stiffness of the resulting spring will be the same as the stiffness of the original spring.

Answer: A) Same

Case 2: When the two halves of the original spring are joined in parallel

In this case, the total number of turns in the resulting spring will be the same as the original number of turns, but the wire diameter will be doubled.

Therefore, the stiffness of the resulting spring can be calculated as follows:

Stiffness ∝ (Number of turns)⁴/(Wire diameter)³

Stiffness of the resulting spring ∝ (Original number of turns)⁴/(2 x Original wire diameter)³

Stiffness of the resulting spring ∝ (1/8) x (Original number of turns)⁴/(Original wire diameter)³

Stiffness of the resulting spring is one-eighth of the stiffness of the original spring.

Answer: D) One-fourth

We know na k1/k2 = n2/n1
solve and find k2=2k1