The principal at the end of the first year is $p$.
$\frac{p\times 10 \times 1}{100} = 132$ $\Rightarrow p = 1320$
Let the actual principal $P$
so, after $1$ year amount will be, $P + \frac{P\times 10 \times 1}{100}$ $\Rightarrow$ $\frac{11P}{100} = 1320$
$P = 1200 /-$
Correct Ans: Option B