Question

If diameter reduces by 50% during wire drawing, area reduces by

• A. 25%
• B. 75%
• C. 50%
• D. 10%

Hint:

For any circular or square section undergoing uniform scaling of its linear dimension \( L \) by a factor \( k \) (where \( L_{\text{new}} = k L_{\text{old}} \)):
- New Area \( A_{\text{new}} = k^2 A_{\text{old}} \).
- In this case, \( k = 0.5 \implies k^2 = 0.25 \).
- Area Reduction = \( 1 - k^2 = 1 - 0.25 = 0.75 \rightarrow 75\% \).
This quick scaling trick avoids tedious calculations during exams.

Answer 1

The Correct Option is B

Step 1: Set up the diameters.
Let the starting diameter of the wire be \( d_0 \). Since drawing cuts the diameter by 50%, the new diameter is \( d_1 = 0.5 d_0 \).

Step 2: Compare the areas using the diameter ratio.
Cross sectional area of a round wire scales with the square of its diameter, so
\[ \frac{A_1}{A_0} = \left(\frac{d_1}{d_0}\right)^2 = (0.5)^2 = 0.25 \]
This means the new area is only a quarter of the original area, even though the diameter only dropped by half.

Step 3: Convert this to percentage reduction.
If the remaining area is 25% of the original, the area that has been removed must be
\[ 100\% - 25\% = 75\% \]

Step 4: Final answer.
\[ \boxed{75\% \text{ reduction in area, option (B)}} \]