Question

Masses of three wires are in the ratio $1:3:5$. Their length are in the ratio $5:3:1$. When connected in series with a battery the ratio of heats produced in them will be

• A. $125:15:1$
• B. $1:15:125$
• C. $5:3:1$
• D. $1:3:5$

Answer 1

Answer: (A)

$125:15:1$

The resistance of wire is given by: $R=\frac{\rho l}{A}$
But, $A=\frac{ml}{d}$
where $m$ is mass, $l$ is length and $d$ is the density of material of conductor.

Hence, $R=\frac{\rho l}{\left(\frac{m}{dl}\right)}$ $=\frac{\rho d{l}^2}{m}$
$\Rightarrow R\propto \frac{l^2}{m}$ .......... (1)

Let, $R_1$, $R_2$ and $R_3$ be the resistances of three wires, $l_1$, $l_2$ and $l_3$ be the lengths of three wires and $m_1$, $m_2$ and $m_3$ be the masses of three wires. Therefore,
$R_1\propto \frac{l_1^2}{m_1}$
$R_2\propto \frac{l_2^2}{m_2}$
$R_3\propto \frac{l_3^2}{m_3}$

Also, heat produced in the wire is: $H=I^{2}R$
Wires are connected in series hence, current through them is same.
Therefore,
$H_1\propto R_1$
$H_2\propto R_2$
$H_3\propto R_3$

From above,
$H_1:H_2:H_3=\left(\frac{l_1^2}{m_1}\right):\left(\frac{l_2^2}{m_2}\right):\left(\frac{l_3^2}{m_3}\right)$

$H_1:H_2:H_3=\left(\frac{5^2}{1}\right):\left(\frac{3^2}{3}\right):\left(\frac{1^2}{5}\right)$

$H_1:H_2:H_3=\left(\frac{25}{1}\right):\left(\frac{9}{3}\right):\left(\frac{1}{5}\right)$

$H_1:H_2:H_3=125:15:1$