Ans: B
Note that the maximum absolute error
Note that the maximum absolute error
$\begin{array}{cl}
= & \dfrac{1}{2} \times 1 \\
= & 0.5\text{ cm}
\end{array}$
Therefore, the least possible radius of the sphere
$\begin{array}{cl}
= & 8 – 0.5 \\
= & 7.5 \text{ cm}
\end{array}$
Therefore, the least possible surface of the sphere
$\begin{array}{cl}
= & 4 \pi (7.5)^2 \\
= & 225\pi\text{ cm}^2
\end{array}$
