2008-II-16

Ans: B
Note that the maximum absolute error

$\begin{array}{cl} = & \frac{1}{2} \times 1 \\ = & 0.5 cm \end{array}$

Therefore, the least possible radius of the sphere

$\begin{array}{cl} = & 8 - 0.5 \\ = & 7.5 cm \end{array}$

Therefore, the least possible surface of the sphere

$\begin{array}{cl} = & 4 π (7.5)^{2} \\ = & 225 π {cm}^{2} \end{array}$