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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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	$(D)$ $2:1$

Similar Questions

Above neutral temperature,thermo $e.m.f.$ in a thermocouple

The expression for thermo $e.m.f.$ in a thermocouple is given by the relation $E = 40\theta - \frac{\theta^2}{20}$,where $\theta$ is the temperature difference of two junctions. For this,the neutral temperature will be ................... $^oC$.

The Peltier coefficient for the junction of a pair of metals is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module