A through hole is drilled in an aluminum alloy plate of 15 mm thickness with a drill bit of diameter 10 mm, at a feed of 0.25 mm/rev and a spindle speed of 1200 rpm. If the specific energy required for cutting this material is 0.7 N-m/mm³, the power required for drilling is W (round off to two decimal places).

The correct answer is 274.89 W (or 274.89).

To find the power required for drilling, we can use the relationship between specific energy consumption, the Material Removal Rate (MRR), and power.

1. Identify the given parameters:
- Thickness of the plate, $t=15\text{mm}$
- Diameter of the drill bit, $d=10\text{mm}$
- Feed rate, $f=0.25\text{mm/rev}$
- Spindle speed, $N=1200\text{rpm}$
- Specific cutting energy, $e_s=0.7{\text{N-m/mm}}^3=0.7{\text{J/mm}}^3$

2. Calculate the Material Removal Rate (MRR):
The rate at which material is removed during drilling is calculated using the cross-sectional area of the drill hole and the axial feed speed:
$\mathrm{MRR}=\frac{\pi \cdot d^2}{4}\cdot f\cdot N$
Substitute the given values into the equation:
$\mathrm{MRR}=\frac{\pi \cdot {10}^2}{4}\cdot 0.25\cdot 1200$
$\mathrm{MRR}=78.5398\cdot 300=23561.94{\text{mm}}^3\text{/min}$
Convert the MRR to units of ${\text{mm}}^3\text{/s}$:
$\mathrm{MRR}=\frac{23561.94}{60}=392.70{\text{mm}}^3\text{/s}$

3. Calculate the Power Required:
The power required is the product of the specific energy consumption and the volumetric material removal rate:
$\mathrm{Power\; (P)}=e_s\cdot \mathrm{MRR}$
$P=0.7{\text{J/mm}}^3\cdot 392.70{\text{mm}}^3\text{/s}$
$P=274.89\text{J/s}=274.89\text{W}$