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**Hint**Connect the resistors in series. We know that when 2 resistors are in series, then the overall resistance is the sum of their individual resistance. After finding the total resistance value, find the heat dissipated value using Joule’s law.

**Complete Step By Step Solution**

It is given that two resistors of 60 ohms and 40 ohms are connected in series across a 200V AC supply. Now, in a series circuit, the overall resistance present in the circuit is equal to the sum of individual resistance values of the resistors. Hence,

\[{R_{eff}} = {R_1} + {R_2}\]

\[ \Rightarrow {R_{eff}} = 60\Omega + 40\Omega = 100\Omega \]

Now applying Joule’s law of heating for the given circuit , we get

\[H = {I^2}Rt\],where I is the current flowing in the circuit, R is the overall resistance and t is the time period of action or current flowing in the circuit.

We don’t know the value, but using ohm’s law we get, \[I = \dfrac{V}{R}\]. Using this on the above joule’s law equation we get,

\[H = {(\dfrac{V}{R})^2}Rt\]

Substituting the known values on the above equation we get,

\[ \Rightarrow H = {(\dfrac{{220}}{{10}})^2}10 \times 30\] (Since it is given that the combination runs for half a minute)

\[ \Rightarrow H = 484 \times 10 \times 30\]

\[ \Rightarrow H = 145200J\]

**Thus, option (d) is the correct answer for the given question.**

**Note**Joule’s law of heating determines the rate at which circuit resistance converts electrical energy to heat energy. It states that when a current ‘I’ passes through a resistor of resistance ‘R’ for a specified time period ‘t’ , the head dissipated is equal to the product of overall resistance, time period of current flow and square of current flowing. Joule’s law is useful to check on the temperature produced by an electrical system and design accordingly to produce less heat.

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