Question:
A wire extends by $1$ mm under $100$ N. Extension under $300$ N is:
A wire extends by $1$ mm under $100$ N. Extension under $300$ N is:
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For linear relationships like Hooke's Law, if the force becomes 'n' times, the extension also becomes 'n' times. Here, since $300$ N is $3$ times $100$ N, the extension is simply $3 \times 1$ mm = $3$ mm.
Updated On: May 20, 2026
- $1$ mm
- $2$ mm
- $3$ mm
- $4$ mm
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The Correct Option is C
Solution and Explanation
Concept:
The problem is based on Hooke's Law, which states that within the elastic limit, the extension ($\Delta L$) of a material is directly proportional to the force ($F$) applied to it.
\[ F \propto \Delta L \quad \Rightarrow \quad F = k\Delta L \]
where $k$ is the force constant of the wire.
Step 1: Establish the relationship between Force and Extension.
Since the same wire is used, the proportionality remains constant: \[ \frac{F_1}{\Delta L_1} = \frac{F_2}{\Delta L_2} \]
Step 2: Substitute the given values.
From the question:
• Initial force ($F_1$) = $100$ N
• Initial extension ($\Delta L_1$) = $1$ mm
• Final force ($F_2$) = $300$ N Plugging these into the ratio: \[ \frac{100}{1} = \frac{300}{\Delta L_2} \]
Step 3: Solve for the new extension.
\[ \Delta L_2 = \frac{300 \times 1}{100} \] \[ \Delta L_2 = 3 \text{ mm} \]
Step 1: Establish the relationship between Force and Extension.
Since the same wire is used, the proportionality remains constant: \[ \frac{F_1}{\Delta L_1} = \frac{F_2}{\Delta L_2} \]
Step 2: Substitute the given values.
From the question:
• Initial force ($F_1$) = $100$ N
• Initial extension ($\Delta L_1$) = $1$ mm
• Final force ($F_2$) = $300$ N Plugging these into the ratio: \[ \frac{100}{1} = \frac{300}{\Delta L_2} \]
Step 3: Solve for the new extension.
\[ \Delta L_2 = \frac{300 \times 1}{100} \] \[ \Delta L_2 = 3 \text{ mm} \]
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