Question:
A steel wire of initial length 149 cm extends by 0.5 cm when a tension of 50 N is applied along its length. If an additional tension of 100 N is applied to the same wire, then the final length of the wire is
A steel wire of initial length 149 cm extends by 0.5 cm when a tension of 50 N is applied along its length. If an additional tension of 100 N is applied to the same wire, then the final length of the wire is
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Within elastic limit:
\[
\frac{\Delta L_1}{\Delta L_2}
=
\frac{F_1}{F_2}
\]
Updated On: Jun 17, 2026
- $150.5\text{ cm}$
- $150\text{ cm}$
- $151\text{ cm}$
- $151.5\text{ cm}$
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The Correct Option is B
Solution and Explanation
Concept:
According to Hooke's law, extension is directly proportional to the applied force within the elastic limit.
\[
\Delta L \propto F
\]
Step 1: Determine extension produced by 150 N force.
Given, \[ 50N \rightarrow 0.5\text{ cm} \] Therefore, \[ 150N \rightarrow \frac{150}{50}\times0.5 \] \[ =1.5\text{ cm} \]
Step 2: Calculate final length.
Initial length \[ L_0=149\text{ cm} \] Final length \[ L=149+1.5 \] \[ L=150.5\text{ cm} \] Hence, \[ \boxed{150.5\text{ cm}} \]
Step 1: Determine extension produced by 150 N force.
Given, \[ 50N \rightarrow 0.5\text{ cm} \] Therefore, \[ 150N \rightarrow \frac{150}{50}\times0.5 \] \[ =1.5\text{ cm} \]
Step 2: Calculate final length.
Initial length \[ L_0=149\text{ cm} \] Final length \[ L=149+1.5 \] \[ L=150.5\text{ cm} \] Hence, \[ \boxed{150.5\text{ cm}} \]
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