Two solid cylinders of equal diameter have different heights. They are compressed plastically bya pair of rigid dies to create the same percentage reduction in their respective heights. Considerthat the die-workpiece interface friction is negligible. The ratio of the final diameter of theshorter cylinder to that of the longer cylinder is __________(Important - Enter only the numerical value in the answer)Correct answer is '1'. Can you explain this answer?

Question

Answer

Explanation:

To solve this problem, let's consider the initial and final dimensions of the two cylinders.

Let:
- D = diameter of the cylinders
- h1 = initial height of the shorter cylinder
- h2 = initial height of the longer cylinder
- h1' = final height of the shorter cylinder
- h2' = final height of the longer cylinder

Since both cylinders are compressed plastically by the same percentage reduction in their heights, we can set up the following equation:

(h1 - h1') / h1 = (h2 - h2') / h2

Step 1: Finding the final height of the longer cylinder
Since both cylinders have the same initial diameter, the percentage reduction in height is the same for both. Therefore, we can write:

(h1 - h1') / h1 = (h2 - h2') / h2

Simplifying this equation, we get:

h1' / h1 = h2' / h2

Cross-multiplying, we get:

h1' * h2 = h1 * h2'

Step 2: Finding the ratio of the final diameters
The initial and final volumes of the cylinders are the same since the material is incompressible. Therefore, we can write:

π * (D/2)^2 * h1 = π * (D/2)^2 * h1'

Cancelling out the common terms, we get:

h1 = h1'

Similarly, we can write:

π * (D/2)^2 * h2 = π * (D/2 * r)^2 * h2'

Cancelling out the common terms and simplifying, we get:

(D/2)^2 * h2 = (D/2 * r)^2 * h2'

D^2 * h2 = (D * r)^2 * h2'

Dividing both sides by h2 and taking the square root, we get:

D = D * r

Simplifying, we get:

1 = r

Therefore, the ratio of the final diameter of the shorter cylinder to that of the longer cylinder is 1.

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