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 ________________ mm (round off to one decimal place).

The correct answer is 3883.5 (or 3883.49 before rounding).

Step 1: Understand the Principle of Volume Constancy
In a metal rolling operation, the volume of the slab remains constant because the material undergoes plastic deformation without any loss or gain of mass/volume. Therefore, we can write:

$$V_i=V_f$$

where $V_i$ is the initial volume and $V_f$ is the final volume of the metal slab.

Step 2: Identify the Given Initial Dimensions
Let the initial thickness, width, and length be denoted as $t_i$, $W_i$, and $L_i$, respectively:
Initial thickness, _{$t_i=50\text{mm}$}
Initial width, _{$W_i=250\text{mm}$}
Initial length, _{$L_i=3600\text{mm}$}

Step 3: Determine the Final Dimensions after Rolling
The thickness reduces by 10%. Thus, the final thickness ($t_f$) is:

$$t_f=t_i\times (1-0.10)=50\times 0.90=45\text{mm}$$

The width increases by 3%. Thus, the final width ($W_f$) is:

$$W_f=W_i\times (1+0.03)=250\times 1.03=257.5\text{mm}$$

Step 4: Calculate the Final Length
Using the volume conservation relation, $L_i\times W_i\times t_i=L_f\times W_f\times t_f$:

$$3600\times 250\times 50=L_f\times 257.5\times 45$$

$$\mathrm{45,000,000}=L_f\times \mathrm{11,587.5}$$

Solving for $L_f$:

$$L_f=\frac{\mathrm{45,000,000}}{\mathrm{11,587.5}}$$

Rounding off to one decimal place, the length of the rolled slab is 3883.5 mm.