Match the List-I with the List-II from the co | vedclass

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(B) The heat produced in a resistor is given by $H = I^2Rt$ or $H = (V^2/R)t$.$(I)$ Same resistance $(R)$, same current $(I)$: $H_1/H_2 = (I^2Rt)/(I^2

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$I-C, II-D, III-C, IV-D$

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(B) The heat produced in a resistor is given by $H = I^2Rt$ or $H = (V^2/R)t$.$(I)$ Same resistance $(R)$, same current $(I)$: $H_1/H_2 = (I^2Rt)/(I^2Rt) = 1:1$. Matches $(C)$.$(II)$ Resistances $R$ and $2R$ in series, same $P.D.$ $(V)$ applied across the combination: The current $I$ is same through both. $H_1/H_2 = (I^2Rt)/(I^2(2R)t) = 1:2$. Matches $(A)$. Wait, the question states same $P.D.$ applied across the combination, so $I$ is same. If $P.D.$ is applied across each, it would be different. Given the context, $I$ is constant in series, so $H \propto R$. Thus $1:2$. Matches $(A)$.$(III)$ Same resistance, same current: $H_1/H_2 = 1:1$. Matches $(C)$.$(IV)$ Resistances $R_1:R_2 = 1:2$ in parallel, same $P.D.$ $(V)$: $H = (V^2/R)t$, so $H \propto 1/R$. $H_1/H_2 = R_2/R_1 = 2/1 = 2:1$. Matches $(D)$.Correct mapping: $I-C, II-A, III-C, IV-D$. Since this is not an option, re-evaluating $(II)$: If $P.D.$ is applied across each, $H_1/H_2 = (V^2/R)/(V^2/2R) = 2:1$. If $I$ is same, $1:2$. Given the options, $(b)$ is the intended answer.

List-$I$	List-$II$
$(I)$ Two wires of same resistance are connected in series and same current is passed through them	$(A)$ $1:2$
$(II)$ Two wires of resistance $R$ and $2R$ $\Omega$ are connected in series and same $P.D.$ is applied across them	$(B)$ $4:1$
$(III)$ Two wires of same resistance are connected in parallel and same current is flowing through them	$(C)$ $1:1$
$(IV)$ Two wires of resistances in the ratio $1:2$ are connected in parallel and same $P.D.$ is applied across them	$(D)$ $2:1$

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