Correct Answer: Rs. $1200$

Solution : Given: Rate of interest = 10% p.a
Compound Interest (C.I) for the second year = Rs. 132
Formula:
$CI = P[(1+\frac{R}{100})^{n}-1]$, where $P$ is the principal, $R$ is the rate of interest, and $n$ is the number of years.
Solution:
C.I of 2nd year – C.I of 1st year = Rs. 132
$P[(1+\frac{10}{100})^{2}–1]–P[(1+\frac{10}{100})^{1}–1]=132$
⇒ $P[\frac{121}{100}–1]–P[1+\frac{1}{10}–1]=132$
⇒ $P[\frac{121}{100}–1]–P[\frac{11}{10}–1]=132$
⇒ $\frac{21P}{100}–\frac{P}{10}=132$
⇒ $\frac{21P–10P}{100}=132$
⇒ $11P=13200$
⇒ $P=1200$
Hence, the correct answer is Rs. $1200$.