Q1: An ingot is hot forged to a 60% reduction in cross-section area. The percentage reduction in the volume for the above process is

ANS:A - 0

To calculate the percentage reduction in volume when an ingot is hot forged to a 60% reduction in cross-sectional area, we'll use the relationship between cross-sectional area and volume in a solid object. The volume of a solid object is directly proportional to the square of its linear dimensions. Therefore, if the cross-sectional area is reduced by a certain percentage, the volume will be reduced by a greater percentage. Let's denote the initial cross-sectional area as A0 and the final cross-sectional area after forging as A. Also, let's denote the initial volume as V0 and the final volume after forging as V. Given that the cross-sectional area is reduced to 60% of its initial value, A = 0.60 * A0. Since the volume is directly proportional to the square of the linear dimensions, we have: V = k * A0^2, where k is a constant. Similarly, after forging, the final volume V is related to the final cross-sectional area A by: V = k * A^2. Therefore, we can write: V = (A / A0)^2 * V0. Substituting the given value for A in terms of A0: V = (0.60)^2 * V0. V = 0.36 * V0. Now, to find the percentage reduction in volume: Percentage reduction in volume = [(Initial volume - Final volume) / Initial volume] * 100%. Percentage reduction in volume = [(V0 - 0.36 * V0) / V0] * 100%. Percentage reduction in volume = [0.64 * V0 / V0] * 100%. Percentage reduction in volume = 64%. So, the correct answer is 64%.