Question:
A system consists of two springs connected in series and each having the spring constant \(10\,\text{N m}^{-1}\). The minimum work required to stretch this system by \(1\,\text{cm}\) in erg is
A system consists of two springs connected in series and each having the spring constant \(10\,\text{N m}^{-1}\). The minimum work required to stretch this system by \(1\,\text{cm}\) in erg is
Show Hint
For springs in series,
\[
\frac{1}{k_{\text{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}.
\]
The work done in stretching a spring is
\[
W=\frac{1}{2}kx^2.
\]
Updated On: Jun 18, 2026
- \(1500\)
- \(2000\)
- \(3000\)
- \(2500\)
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The Correct Option is D
Solution and Explanation
Step 1: Find the equivalent spring constant.
For two springs connected in series, \[ \frac{1}{k_{\text{eq}}}=\frac{1}{k_1}+\frac{1}{k_2} \] Here, \[ k_1=10\,\text{N m}^{-1} \] and \[ k_2=10\,\text{N m}^{-1} \] Therefore, \[ \frac{1}{k_{\text{eq}}}=\frac{1}{10}+\frac{1}{10} \] \[ \frac{1}{k_{\text{eq}}}=\frac{2}{10} \] \[ k_{\text{eq}}=5\,\text{N m}^{-1} \]
Step 2: Convert extension into metre.
Given extension is \[ x=1\,\text{cm} \] \[ x=10^{-2}\,\text{m} \]
Step 3: Calculate the work required.
The minimum work required to stretch a spring is stored as elastic potential energy, \[ W=\frac{1}{2}k_{\text{eq}}x^2 \] Substituting the values, \[ W=\frac{1}{2}(5)(10^{-2})^2 \] \[ W=\frac{5}{2}\times 10^{-4} \] \[ W=2.5\times 10^{-4}\,\text{J} \]
Step 4: Convert joule into erg.
We know, \[ 1\,\text{J}=10^7\,\text{erg} \] Therefore, \[ W=2.5\times 10^{-4}\times 10^7\,\text{erg} \] \[ W=2.5\times 10^3\,\text{erg} \] \[ W=2500\,\text{erg} \]
Step 5: Final conclusion.
Therefore, the minimum work required is \[ \boxed{2500\,\text{erg}} \]
For two springs connected in series, \[ \frac{1}{k_{\text{eq}}}=\frac{1}{k_1}+\frac{1}{k_2} \] Here, \[ k_1=10\,\text{N m}^{-1} \] and \[ k_2=10\,\text{N m}^{-1} \] Therefore, \[ \frac{1}{k_{\text{eq}}}=\frac{1}{10}+\frac{1}{10} \] \[ \frac{1}{k_{\text{eq}}}=\frac{2}{10} \] \[ k_{\text{eq}}=5\,\text{N m}^{-1} \]
Step 2: Convert extension into metre.
Given extension is \[ x=1\,\text{cm} \] \[ x=10^{-2}\,\text{m} \]
Step 3: Calculate the work required.
The minimum work required to stretch a spring is stored as elastic potential energy, \[ W=\frac{1}{2}k_{\text{eq}}x^2 \] Substituting the values, \[ W=\frac{1}{2}(5)(10^{-2})^2 \] \[ W=\frac{5}{2}\times 10^{-4} \] \[ W=2.5\times 10^{-4}\,\text{J} \]
Step 4: Convert joule into erg.
We know, \[ 1\,\text{J}=10^7\,\text{erg} \] Therefore, \[ W=2.5\times 10^{-4}\times 10^7\,\text{erg} \] \[ W=2.5\times 10^3\,\text{erg} \] \[ W=2500\,\text{erg} \]
Step 5: Final conclusion.
Therefore, the minimum work required is \[ \boxed{2500\,\text{erg}} \]
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