The same mass of copper is drawn into two wir | vedclass

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(D) The heat produced in a wire is given by $H = I^2 R t$. Since the wires are connected in series, the current $I$ and time $t$ are the same for both

Vedclass · Accepted Answer

$16:1$

Vedclass · Answer

(D) The heat produced in a wire is given by $H = I^2 R t$. Since the wires are connected in series, the current $I$ and time $t$ are the same for both. Resistance $R = ho \frac{l}{A}$. Since the mass $m = 	ext{density} 	imes 	ext{volume} = ho_d 	imes A 	imes l$ is constant, $l = \frac{m}{ho_d A}$. Substituting this into the resistance formula: $R = ho \frac{m}{ho_d A^2} \propto \frac{1}{A^2}$. Since the area $A = \pi r^2$, we have $R \propto \frac{1}{(r^2)^2} = \frac{1}{r^4}$. Therefore, $H \propto \frac{1}{r^4}$. For the two wires with radii $r_1 = 1 \, mm$ and $r_2 = 2 \, mm$, the ratio of heat produced is: $\frac{H_1}{H_2} = \left( \frac{r_2}{r_1} ight)^4 = \left( \frac{2}{1} ight)^4 = \frac{16}{1}$.

The same mass of copper is drawn into two wires $1 \, mm$ and $2 \, mm$ thick. Two wires are connected in series and current is passed through them. The ratio of heat produced in the wires is:

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