Answer
A)length of the beam after shrinking should be 12.02 meters
B)Water content should be 234.38 \frac{kg}{m^{3}}
Work Step by Step
A) Using the given formula S=\frac{0.032(250)-2.5}{10000}=
shrinking factor->0.00055
length of the beam after shrinking=12.025\times 0.99945 \approx 12.02
B)
0.00050=\frac{0.032w-2.5}{10000}
-> 5=0.032w-2.5
->7.5=0.032w
->w=234.38
Water content should be 234.38 \frac{kg}{m^{3}}
