Two heater wires of equal length are first connected in series and then in parallel The ratio of heat produced in the two cases is (A) \(2: 1\) (B) \(1: 2\) (C) \(4: 1\) (D) \(1: 4\)

Let the resistance of each heater wire be R.

The equivalent resistance when the heater wires are connected in series is given by: \(R_{series} = R_1 + R_2 = R + R = 2R\)

The equivalent resistance when the heater wires are connected in parallel is given by: \(\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R} + \frac{1}{R}\) \(R_{parallel} = \frac{R}{2}\)

As we know, the power in a resistor circuit is given by: \(P = I^2 R\) In both cases, the voltage applied across the heaters will remain the same. So, let the current passing through series-connected heaters be \(I_{series}\), and through parallel-connected heaters be \(I_{parallel}\). Power in the series case can be expressed as: \(P_{series} = I_{series}^2 R_{series} = I_{series}^2 (2R)\) Power in the parallel case can be expressed as: \(P_{parallel} = I_{parallel}^2 R_{parallel} = I_{parallel}^2 (\frac{R}{2})\)

The ratio of heat produced in the series and parallel cases can be given by: \(\frac{P_{series}}{P_{parallel}} = \frac{I_{series}^2 (2R)}{I_{parallel}^2 (\frac{R}{2})}\) Since both heaters are connected across the same voltage source, we know that: \(V = I_{series} R_{series} = I_{parallel} R_{parallel}\) Using this relation, we have: \(\frac{I_{series}}{I_{parallel}} = \frac{R_{parallel}}{R_{series}} \) Now we can substitute and simplify: \(\frac{P_{series}}{P_{parallel}} = \frac{(2R)(\frac{R_{parallel}}{R_{series}})^2}{\frac{R}{2}}\) \(\frac{P_{series}}{P_{parallel}} = \frac{(2R)(\frac{\frac{R}{2}}{2R})^2}{\frac{R}{2}}\) \(\frac{P_{series}}{P_{parallel}} = \frac{1}{4}\) The ratio of heat produced in the two cases is \(1:4\), which corresponds to option (D). So, the correct answer is (D) \(1: 4\).