Question

The thickness, width and length of a metal slab are 50 mm, 250 mm and 3600 mm, respectively. A rolling operation on this slab reduces the thickness by 10% and increases the width by 3%. The length of the rolled slab is \(\underline{\hspace{2cm}}\) mm (round off to one decimal place).

Hint:

Volume conservation in rolling operations allows us to calculate the final length of the material using the change in thickness and width.

Answer 1

Correct Answer: 3880 - 3886

The volume of the slab before rolling is given by: \[ V_{\text{initial}} = \text{thickness} \times \text{width} \times \text{length} \] Substituting the values: \[ V_{\text{initial}} = 50 \times 250 \times 3600 = 45,000,000 \ \text{mm}^3 \] After the rolling operation, the thickness is reduced by 10% and the width increases by 3%. Therefore:
- New thickness \( = 50 \times (1 - 0.10) = 45 \ \text{mm} \),
- New width \( = 250 \times (1 + 0.03) = 257.5 \ \text{mm} \).
Let \( L_{\text{final}} \) be the final length of the slab. The volume remains constant, so: \[ V_{\text{initial}} = V_{\text{final}} \] \[ 45,000,000 = 45 \times 257.5 \times L_{\text{final}} \] Solving for \( L_{\text{final}} \): \[ L_{\text{final}} = \frac{45,000,000}{45 \times 257.5} = 3886 \ \text{mm} \] Thus, the length of the rolled slab is: \[ \boxed{3886.0 \ \text{mm}} \]