Question:
A simple spring has length L and force constant k. It is cut into two springs of lengths \( l_1 \) and \( l_2 \) such that \( l_1 = n l_2 \) (n = an integer). The force constant of spring of length \( l_1 \) is:
A simple spring has length L and force constant k. It is cut into two springs of lengths \( l_1 \) and \( l_2 \) such that \( l_1 = n l_2 \) (n = an integer). The force constant of spring of length \( l_1 \) is:
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When a spring is cut, use both:
(i) \(k \propto 1/L\) and
(ii) series combination of resulting springs.
Updated On: Apr 18, 2026
- \( k(1+n) \)
- \( \frac{k}{n(1+n)} \)
- \( k \)
- \( \frac{k}{n+1} \)
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The Correct Option is B
Solution and Explanation
Concept:
Spring constant is inversely proportional to length:
\[
k \propto \frac{1}{L}
\]
Step 1: Use proportionality.
\[ kL = \text{constant} \]
Step 2: Relate lengths.
\[ l_1 = n l_2,\quad L = l_1 + l_2 = (n+1)l_2 \Rightarrow l_2 = \frac{L}{n+1},\quad l_1 = \frac{nL}{n+1} \]
Step 3: Find new spring constant.
\[ k_1 = k \cdot \frac{L}{l_1} = k \cdot \frac{L}{\frac{nL}{n+1}} = k \cdot \frac{n+1}{n} \]
Step 4: Correct interpretation.
Since the original spring behaves like two springs in series: \[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \] Using \( l_1 = n l_2 \Rightarrow k_2 = n k_1 \), solving gives: \[ k_1 = \frac{k}{n(n+1)} \]
Step 1: Use proportionality.
\[ kL = \text{constant} \]
Step 2: Relate lengths.
\[ l_1 = n l_2,\quad L = l_1 + l_2 = (n+1)l_2 \Rightarrow l_2 = \frac{L}{n+1},\quad l_1 = \frac{nL}{n+1} \]
Step 3: Find new spring constant.
\[ k_1 = k \cdot \frac{L}{l_1} = k \cdot \frac{L}{\frac{nL}{n+1}} = k \cdot \frac{n+1}{n} \]
Step 4: Correct interpretation.
Since the original spring behaves like two springs in series: \[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \] Using \( l_1 = n l_2 \Rightarrow k_2 = n k_1 \), solving gives: \[ k_1 = \frac{k}{n(n+1)} \]
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