A player tosses two coins. He wins Rs. 10, if 2 heads appears, Rs. 5, if one head appear and Rs. 2 if no head appears. Then variance of winning amount is

Expected value and variance for the two-coin toss game

In a game where two fair coins are tossed, the possible outcomes are $HH$, $HT$, $TH$, and $TT$, each with probability $\frac{1}{4}$.
The winning amount $X$ is $10$ for $HH$ ($P(X=10)=1/4$), $5$ for $HT$ or $TH$ ($P(X=5)=1/2$), and $2$ for $TT$ ($P(X=2)=1/4$).

The expected value is $E(X) = \sum x \cdot P(X=x) = 10 \cdot \frac{1}{4} + 5 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} = 5.5$.

$E(X^2) = 100 \cdot \frac{1}{4} + 25 \cdot \frac{1}{2} + 4 \cdot \frac{1}{4} = 38.5$.

Variance is $\text{Var}(X) = E(X^2) - [E(X)]^2 = 38.5 - (5.5)^2 = 8.25$.

Hence, the variance is $\boxed{\text{C}}$.