Question

Four wires of the same diameter are connected, in turn, between two points maintained at a constant potential difference. Their resistivities and lengths are; $\rho $ and $L$ (wire1), $1.2\rho $ and $1.2L$ (wire2), $0.9\rho $ and $0.9L$ (wire3) and $\rho $ and $1.5L$ (wire4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first.

$\text{A}\text{. }4 > 3 > 1 > 2$
$\text{B}\text{. }4 > 2 > 1 > 3$
$\text{C}\text{. 1} > 2 > 3 > 4$
$\text{D}\text{. 3} > 1 > 2 > 4$

Answer 1

Hint: For calculating the value of heat dissipation across each wire, we need to calculate the resistance of the wires. By using joule’s law of heating we can calculate the value of energy dissipated in the form of heat.

Formula used:

$R=\dfrac{\rho L}{A}$
$H=\dfrac{{{V}^{2}}}{R}$

Complete step by step answer:

Joule’s law of heating says that when a current $I$ passes through a conductor of resistance $R$ for time $t$, then the heat developed in the conductor will be given as $H={{I}^{2}}Rt$

Rate of heat dissipation $=\dfrac{H}{t}={{I}^{2}}R$

By Ohm’s law, \[I=\dfrac{V}{R}\]

Therefore, Rate of heat dissipation\[=\dfrac{{{V}^{2}}}{R}\]

Resistance of wire is given as, $R=\dfrac{\rho L}{A}$

Where,
$\rho $ is the resistivity of the wire
$L$ is the length of the wire
$A$ is the area of cross section of wire

Resistivity of wire will be $\rho =\dfrac{RA}{L}$

For finding the rate of heat energy dissipated, we will use the formula,

$H=\dfrac{{{V}^{2}}}{R}$

Where,
$V$ is the potential difference across two fixed points
$R$ is the resistance of the wire
Or, $H=\dfrac{{{V}^{2}}A}{\rho L}$
According to the connection of wires, voltage drop $V$ across all the four wires will be equal.

For wire 1,

Rate of heat dissipation ${{H}_{1}}=\dfrac{{{V}^{2}}A}{\rho L}$

For wire 2,

Rate of heat dissipation ${{H}_{2}}=\dfrac{{{V}^{2}}A}{(1.2\rho )\times (1.2L)}=\dfrac{0.694{{V}^{2}}A}{\rho L}$
${{H}_{2}}=0.694{{H}_{1}}$

For wire 3,

Rate of heat dissipation ${{H}_{3}}=\dfrac{{{V}^{2}}A}{(0.9\rho )\times (0.9L)}=\dfrac{1.23{{V}^{2}}A}{\rho L}$
${{H}_{3}}=1.23{{H}_{1}}$

For wire 4,

Rate of heat dissipation ${{H}_{4}}=\dfrac{{{V}^{2}}A}{(\rho )\times (1.5L)}=\dfrac{0.666{{V}^{2}}A}{\rho L}$
${{H}_{4}}=0.666{{H}_{1}}$

Comparing the values of ${{H}_{1}},{{H}_{2}},{{H}_{3}},{{H}_{4}}$
We get $ {{H}_{3}}>{{H}_{1}}>{{H}_{2}}>{{H}_{4}}$

Therefore, Rate of heat dissipation in wires is given as, $\text{3}>1>2>4$

Hence, the correct option is D.

Note: In the above connection the wires were connected in turn, between two points maintained at a constant potential difference, meaning that the combination was parallel and therefore we took the equal value of voltage drop across each wire.