To solve this problem, we need to determine the ratio of thermal energy developed in two resistors connected in parallel.$\\$The thermal energy $(H)$ developed in a resistor is given by the formula:$\\ H = I^2 R t \\$where $I$ is the current through the resistor, $R$ is the resistance, and $t$ is the time.$\\$For resistors in parallel, the voltage across each resistor is the same. Let $V$ be this voltage.$\\$The current through each resistor is given by Ohm's law:$\\ I = \frac{V}{R} \\$Therefore, the thermal energy developed in the $100 \Omega$ resistor is:$\\ H_1 = \left(\frac{V}{100}\right)^2 \cdot 100 \cdot t = \frac{V^2}{100} \cdot t \\$The thermal energy developed in the $200 \Omega$ resistor is:$\\ H_2 = \left(\frac{V}{200}\right)^2 \cdot 200 \cdot t = \frac{V^2}{200} \cdot t \\$Now, we find the ratio of $H_1$ to $H_2: \\ \frac{H_1}{H_2} = \frac{\frac{V^2}{100} \cdot t}{\frac{V^2}{200} \cdot t} \\$Simplifying the expression:$\\ \frac{H_1}{H_2} = \frac{\frac{1}{100}}{\frac{1}{200}} = \frac{200}{100} = 2 \\$Therefore, the ratio of thermal energy developed in the $100 \Omega$ resistor to that in the $200 \Omega$ resistor is $2:1. \\$This corresponds to Option 2.