The reactance offered by a capacitor to alternating current is inversely proportional to the frequency. Therefore, if the frequency is increased to 100 Hz, the reactance will decrease to 10 ohms.
The reactance of a capacitor is given by the following formula:
$X_C = \frac{1}{2\pi f C}$
where $f$ is the frequency in hertz and $C$ is the capacitance in farads.
In this case, we are given that $f = 50$ Hz and $X_C = 20$ ohms. Substituting these values into the formula, we get:
$20 = \frac{1}{2\pi (50 \text{ Hz}) C}$
Solving for $C$, we get:
$C = \frac{1}{2\pi (50 \text{ Hz}) (20 \text{ ohms})} = 10^{-6} \text{ farads}$
If the frequency is increased to 100 Hz, the reactance will be given by:
$X_C = \frac{1}{2\pi (100 \text{ Hz}) (10^{-6} \text{ farads})} = 10 \text{ ohms}$