The required centre-to-centre spacing of 10 mm diameter bars is 150 mm to resist the design moment in a concrete slab. Instead of 10 mm diameter bars, if 12 mm diameter bars of the same grade are used, the required centre-to-centre spacing (in mm) to resist the same design moment becomes_________(rounded of to the nearest integer).

The correct option is 216.

To resist the same design moment in a concrete slab, the total area of steel reinforcement per unit width of the slab must remain the same. The area of steel reinforcement per meter width of a slab is given by the formula:
$$A_{\mathrm{st}}=\frac{a_{\phi }\times 1000}{S}$$
where:
• $a_{\phi }$ is the cross-sectional area of a single bar of diameter $d$, calculated as $a_{\phi }=\frac{\pi }{4}{d}^2$
• $S$ is the center-to-center spacing of the bars in mm.

Since the same design moment is resisted using the same grade of steel, the required area of steel $A_{\mathrm{st}}$ must be kept constant. Therefore, we can set up the relationship:
$$\frac{a_{\phi 1}}{S_1}=\frac{a_{\phi 2}}{S_2}$$
Substituting the area formula $a_{\phi }=\frac{\pi }{4}{d}^2$ into the relation gives:
$$\frac{{d_1}^2}{S_1}=\frac{{d_2}^2}{S_2}$$

Given data:
• Initial bar diameter, $d_1=10\text{mm}$
• Initial spacing, $S_1=150\text{mm}$
• New bar diameter, $d_2=12\text{mm}$

We need to find the new spacing, $S_2$. Rearranging the formula to solve for $S_2$:
$$S_2=S_1\times {\left(\frac{d_2}{d_1}\right)}^2$$
Substituting the given values:
$$S_2=150\times {\left(\frac{12}{10}\right)}^2$$
$$S_2=150\times 1.44=216\text{mm}$$

Thus, the required center-to-center spacing of the 12 mm diameter bars is exactly 216 mm.