Question

The same mass of copper is drawn into two wires $A$ and $B$ of radii $r$ and $3r$ respectively. They are connected in series, and electric current is passed. The ratio of the heat proiduced in $A$ and $B$ is :

• A. $1:9$
• B. $1:81$
• C. $81:1$
• D. $9:1$

Answer 1

The correct option is C $81:1$

Let $R_A$ and $R_B$ are resistances
$R_A=\rho \frac{l_A}{A_A}{R}_B=\rho \frac{l_B}{A_B}$
$\frac{R_A}{R_B}=\frac{l_A}{l_B}\times \frac{A_B}{A_A}={\left(\frac{A_B}{A_A}\right)}^2$
We know,
Mass=Volume $\times$ Density$\left({\rho }_{Cu}\right)$
$[\therefore (l_A{A}_A)\times {\rho }_{Cu}=(l_B{A}_B)\times {\rho }_{Cu}\Rightarrow \frac{A_B}{A_A}=\frac{l_A}{l_B}]$
$=\frac{(\pi \times 9r^2{)}^2}{(\pi r^2{)}^2}=\frac{81}{1}$
In series, Ratio of heat produced $H_A:H_B=R_A:R_B=81:1$