Question

A spring of force constant $15\,\text{N/m}$ is cut into two pieces. If the ratio of their lengths is $1:3$, then the force constant of the smaller piece is ___ $\text{N/m}$.

• A. $60$
• B. $45$
• C. $20$
• D. $15$

Hint:

When a spring is cut, its material properties remain the same; force constant depends on how extension is defined for the piece.

Answer 1

The Correct Option is D

To solve this problem, we need to understand how the force constant of a spring changes when the spring is cut into parts.

The force constant (or spring constant) of a spring is inversely proportional to its length when the material and cross-section are constant. Hence, if a spring of length \(L\) with force constant \(k\) is cut into two parts of lengths \(L_1\) and \(L_2\), the new force constants, \(k_1\) and \(k_2\), can be determined as follows:

• The original spring constant: \(k = 15 \, \text{N/m}\).
• The ratio of the lengths of the parts: \(1:3\).
  • If the total original length \(L = L_1 + L_2\), let \(L_1 = x\) and \(L_2 = 3x\).

Step 1: Express total length in terms of proportional parts.

Since the ratio of lengths is \(1:3\), we can write:

\(L = L_1 + L_2 = x + 3x = 4x\)

Step 2: Calculate the force constant of the smaller piece.

The force constant of the new segment, \(k_1\) (for the length \(L_1 = x\)), is given by:

\(k_1 = \frac{k \cdot L}{L_1} = \frac{15 \cdot 4x}{x} = 60 \, \text{N/m}\)

Hence, the force constant of the smaller piece is 60 N/m.

Conclusion: The correct answer is \(60 \, \text{N/m}\).