Question

A wire of length L and three identical cells of negligible internal resistance are connected in series. Due to current, the temperature of the wire is raised by $\Delta T$ in time t. N identical cells are now connected in series with a wire of length 2L of the same material and cross-section. The temperature is raised by the same amount $\Delta T$ at the same time t. The value of N is:
A) 6
B) 4
C) 8
D) 9

Answer 1

The concept to be used in this question to arrive at the answer is Joule’s law of heating. Joule’s law of heating gives us the relation between the heat produced to the electric current and the resistance.
Heat produced due to electric current, $H = {I^2}Rt$
where $I$ is the current produced, R is the resistance and t is the time elapsed for the flow of the current through the resistance.

Complete step by step solution:
Let us consider a wire of length L connected to 3 identical cells of negligible internal resistance.
If the internal resistance is negligible, the terminal voltage resulting from the 3 cells will be equal to 3V.
By applying Ohm’s law:
$R = \dfrac{V}{I}$
The current in the circuit, $I = \dfrac{{3V}}{R}$
Applying the Joule’s law of heating, we have –
Heat energy, $H = {I^2}Rt$
Substituting the current in the equation, we get –
$H = {\left( {\dfrac{{3V}}{R}} \right)^2}Rt$
Solving,
$H = \dfrac{{9{V^2}t}}{R}$

This heat energy produced in the wire is responsible for rising its temperature by $\Delta {T_1}$.
$H = ms\Delta {T_1}$
where m is the mass of the wire and s is the specific heat capacity of the wire.
Hence equating them, we have –
$\dfrac{{9{V^2}t}}{R} = ms\Delta {T_1}$
$\Delta {T_1} = \dfrac{{9{V^2}t}}{{msR}}$

N identical cells are connected to another wire of the same material but different length. Thus, the terminal voltage will be $nV$.
The resistance of a material depends on its cross-sectional area, length and resistivity as follows:
$R = \dfrac{{\rho l}}{A}$
Resistance of the wire at length L is – $R = \dfrac{{\rho L}}{A}$
Resistance of the wire at length 2L of the same material and cross-sectional area is – $R' = \dfrac{{\rho \left( {2L} \right)}}{A}$
$\dfrac{{R'}}{R} = \dfrac{{\dfrac{{\rho \left( {2L} \right)}}{A}}}{{\dfrac{{\rho L}}{A}}} = 2$
$R' = 2R$
When N identical cells are connected to the wire of resistance $R'$, the heat energy dissipated will be,
$H' = {I'^2}R't$
$H' = {\left( {\dfrac{{nV}}{{R'}}} \right)^2}R't$ $\because I' = \dfrac{{nV}}{{R'}}$
$H' = \dfrac{{{n^2}{V^2}t}}{{R'}}$
This heat energy produced in the wire is responsible for rising its temperature by $\Delta {T_2}$.
$H = 2ms\Delta {T_2}$
where m is the mass of the wire and s is the specific heat capacity of the wire.
Here, the mass gets doubled since the length doubles with the cross-sectional area remaining constant.
Hence equating them, we have:
$\dfrac{{{n^2}{V^2}t}}{{R'}} = 2ms\Delta {T_2}$
$\Delta {T_2} = \dfrac{{{n^2}{V^2}t}}{{2msR'}}$
As per the question, temperature rise $\Delta {T_1}$ and $\Delta {T_2}$ will be equal. Hence, equating them, we get –
$\Delta {T_1} = \Delta {T_2}$
$\dfrac{{9{V^2}t}}{{msR}} = \dfrac{{{n^2}{V^2}t}}{{2msR'}}$
Substituting for $R'$, we get –
$\dfrac{{9{V^2}t}}{{msR}} = \dfrac{{{n^2}{V^2}t}}{{2ms\left( {2R} \right)}}$
Cancelling out the common terms, we arrive at the equation –
$\dfrac{{{n^2}}}{4} = 9$
$\Rightarrow n = \sqrt {36} = 6$
Thus, the number of identical cells connected, $N = 6$.

Hence, the correct option is Option (A).

Note: In this problem, it is explicitly mentioned that the length changes with the cross-sectional area remaining the same. So, we could have directly taken that the resistance is directly proportional to the length and hence, the resistance will double if the length is doubled, right?
This is where the students tend to overstep the line.
Resistance is not just dependent independently on the length or the cross-section but, it is dependent on the ratio of the length and cross-section. So, unless they mention about the cross-section remaining the same, do not make the mistake of assuming it directly, and considering the direct proportionality to length. This tip can prevent you from making mistakes in competitive exams.