Question

The same mass of copper is drawn into wires $2mm$ and $3mm$ thick. The wires are connected in series and current is passed through them. The ratio of heats produced in two wires is

• A. $\frac{9}{4}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{81}{16}$

Answer 1

The correct option is D $\frac{81}{16}$

Let the length of wire $1$ be $L_1\left(3mm\right)$ and wire $2$ be $L_2\left(2mm\right)$

Equating volume of the two wires,
$\pi R_1^2{L}_1=\pi R_2^2{L}_2\frac{L_1}{L_2}=\frac{R_2^2}{R_1^2}=\frac{4}{9}$

Therefore the heat ratio would be,
$\frac{H_1}{H_2}=\frac{V^2{R}_2}{V^2{R}_1}=\frac{\rho L_2{A}_1}{\rho L_1{A}_2}=\frac{L_2{R}_1^2}{L_1{R}_2^2}=\frac{81}{16}$