Pressure for Frozen Water: Solving 6.6*10^(8) N/m^2

[fightboy]

When water freezes into ice it expands in volume by 9.05 percent. Suppose a volume of water is in a household water pipe or a cavity in a rock. If the water freezes, what pressure must be exerted on it to keep its volume from expanding? (If the pipe or rock cannot supply this pressure, the pipe will burst or the rock will split.)
The bulk modulus for ice is 8x10^(9) N/m^(2)
I tried solving this equation by using the formula Δp=-B(ΔV/V_{0).
Which led to the answer Δp= -(8x10^(9) N/m^(2))*(0.0905)= -7.24*10^(8) N/m^(2). This turned out to be incorrect, and the correct answer is 6.6*10^(8) N/m^(2). What did I do wrong? is the volume change not 0.0905? If so what is it? Also how did the final answer turn out to be positive?}

 

[gezibash]

Imagine that the worst case scenario has happened and that water has expanded it's volume by 9.05%. This now is the actual volume of the ice. You're looking for pressure differential that would actually push the ice back to it's original volume. So consider,

$$v_0 = 1.0905 v_p$$
where $v_p$ stands for the pipe volume. Take your ratio as,

$$\frac{\Delta v}{v_0} = \frac{v_p - v_0}{v_0} = \frac{v_p - 1.0905 v_p}{1.0905 v_p}$$
And when you use the formula you showed, it will amount to the correct result I assume.