Question:

When a spring of spring constant $k$ is cut into two pieces whose lengths are $l_1$ and $l_2$, then the ratio of their spring constants $k_1$ and $k_2$ is

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When a spring is cut into smaller sections, the stiffness of each section increases because the force required for unit displacement increases.
Updated On: Mar 5, 2025
  • $\frac{l_2}{l_1}$
  • $\frac{l_1}{l_2}$
  • $\sqrt{l_1 l_2}$
  • $l_1 l_2$
  • $\frac{1}{l_1 l_2}$ 

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The Correct Option is A

Solution and Explanation

Step 1: The spring constant of a spring is inversely proportional to its length when the spring is cut into smaller sections. Mathematically, \[ k' = \frac{k}{l} \] where $k'$ is the spring constant of a smaller piece and $l$ is its length. 
Step 2: When the original spring of constant $k$ is divided into two sections of lengths $l_1$ and $l_2$, their respective spring constants are: \[ k_1 = \frac{k}{l_1}, \quad k_2 = \frac{k}{l_2} \] Step 3: The ratio of $k_1$ to $k_2$ is: \[ \frac{k_1}{k_2} = \frac{\frac{k}{l_1}}{\frac{k}{l_2}} = \frac{l_2}{l_1} \] Step 4: Therefore, the correct answer is (A). 

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