What is the ratio of power dissipated if two electric bulbs are connected in series and the ratio of their resistance is 1:2?

The correct option is 1:2.

Let the two electric bulbs have resistances $R_1$ and $R_2$. The ratio of their resistances is given as:
$\frac{R_1}{R_2}=\frac{1}{2}$

When the two bulbs are connected in series, the same electric current ($I$) flows through both of them.

The power ($P$) dissipated in a resistor of resistance $R$ when a current $I$ passes through it is given by the formula:
$P=I^{2}R$

Therefore, the power dissipated by the first bulb ($P_1$) and the second bulb ($P_2$) are:
$P_1=I^2{R}_1$
$P_2=I^2{R}_2$

Now, let us find the ratio of the power dissipated by the two bulbs:
$\frac{P_1}{P_2}=\frac{I^2{R}_1}{I^2{R}_2}$

Since the current $I$ is the same in a series connection, the term $I^2$ cancels out from both the numerator and the denominator:
$\frac{P_1}{P_2}=\frac{R_1}{R_2}$

Substituting the given ratio of resistances:
$\frac{P_1}{P_2}=\frac{1}{2}$

Thus, the ratio of the power dissipated is 1:2.