1. Given $a, b \in \mathbf{N}, a^{100}$ is a 120-digit number, and $a^{b}$ is a 10-digit number. Then the value of $b$ is ( ).
(A) 7
(B) 8
(C) 9
(D) 10
$-、 1 . B$.
From the given, we have $10^{119} \leqslant a^{100}<10^{120}, 10^{9} \leqslant a^{6}<10^{10}$. Taking the logarithm, we get $\lg 10^{119} \leqslant \lg a^{100}<\lg 10^{120}$, simplifying to $1.19 \leqslant \lg a<1.2$.
Therefore, $1.19 b<b \lg a<1.2 b$.
Also, $9 \leqslant b \lg a<10$, so $b=8$.