Two wires 'A' and 'B' of the same material ha | vedclass

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(D) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.For wires $A$ and $B$,the ratio of resistances is $\frac{R_A}

Vedclass · Accepted Answer

$8:1$

Vedclass · Answer

(D) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.For wires $A$ and $B$,the ratio of resistances is $\frac{R_A}{R_B} = \frac{l_A}{l_B} 	imes \left( \frac{r_B}{r_A} ight)^2$.Given $\frac{l_A}{l_B} = \frac{1}{2}$ and $\frac{r_A}{r_B} = \frac{2}{1}$,so $\frac{r_B}{r_A} = \frac{1}{2}$.Substituting these values: $\frac{R_A}{R_B} = \frac{1}{2} 	imes \left( \frac{1}{2} ight)^2 = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.Since the wires are connected in parallel,the potential difference $V$ across both is the same.The heat produced is given by $H = \frac{V^2}{R} t$.Therefore,the ratio of heat produced is $\frac{H_A}{H_B} = \frac{V^2/R_A}{V^2/R_B} = \frac{R_B}{R_A} = \frac{8}{1}$.

List-$I$	List-$II$
$I$. Joule	$A$. $\text{Henry} \times \text{Amp/sec}$
$II$. Watt	$B$. $\text{Farad} \times \text{Volt}$
$III$. Volt	$C$. $\text{Coulomb} \times \text{Volt}$
$IV$. Coulomb	$D$. $\text{Oersted} \times \text{cm}$
	$E$. $\text{Amp} \times \text{Gauss}$
	$F$. $\text{Amp}^2 \times \text{Ohm}$

Two wires '$A$' and '$B$' of the same material have their lengths in the ratio $1 : 2$ and radii in the ratio $2 : 1$. The two wires are connected in parallel across a battery. The ratio of the heat produced in '$A$' to the heat produced in '$B$' for the same time is

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