Question

A wire extends by $1$ mm under $100$ N. Extension under $300$ N is:

• A. $1$ mm
• B. $2$ mm
• C. $3$ mm
• D. $4$ mm

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem is an application of Hooke's Law, which governs the elastic behavior of materials within their proportionality limit.
Hooke's Law states that the force (\(F\)) needed to extend or compress a spring/wire by some distance (\(\Delta L\)) is linearly proportional to that distance.
Since it is the same wire, the material properties (Young's Modulus), initial length, and area of cross-section are constant.
Step 2: Key Formula or Approach:
The relationship is \(F = k \Delta L\), where \(k\) is the force constant of the wire.
Because the same wire is used in both cases:
\[ \frac{F_1}{\Delta L_1} = \frac{F_2}{\Delta L_2} \]
Step 3: Detailed Explanation:
From the problem description:
- Initial force \(F_1 = 100\) N.
- Initial extension \(\Delta L_1 = 1\) mm.
- Final force \(F_2 = 300\) N.
- We need to find \(\Delta L_2\).
Plugging the values into the proportionality ratio:
\[ \frac{100}{1} = \frac{300}{\Delta L_2} \]
By cross-multiplying:
\[ 100 \times \Delta L_2 = 300 \times 1 \]
\[ \Delta L_2 = \frac{300}{100} \]
\[ \Delta L_2 = 3 \text{ mm} \]
Step 4: Final Answer:
The extension produced under 300 N is 3 mm.