A solid cylinder of diameter 100 mm and height 50 mm is forged between two frictionless flat dies to a height of 25 mm. The percentage change in diameter is
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Since height is halved ($50 \text{ mm}$ to $25 \text{ mm}$), the area must double to keep volume constant.
Thus, $D_2^2 = 2 D_1^2 \implies D_2 = \sqrt{2} D_1 \approx 1.414 D_1$. This immediately gives a $41.4\%$ increase.