Question

A wire of length l and 3 identical cells of negligible internal resistance are connected in series. Due to the current, the temperature of the wire is raised by $\vartriangle \text{T}$ in the time t. A number N of similar cells is now connected in series with a wire of the same material and cross section but length 2l. The temperature of the wire is raised by the same amount $\vartriangle \text{T}$ at the same time t. the values of $\vartriangle \text{l}$ is:
A. 3
B. 2
C. 6
D. 4

Answer 1

The heat produced $\left( {{\text{I}}^{2}}\text{Rt} \right)$ is made equal to Q, which is given by $\text{mc}\vartriangle \text{T}$ . By equating these equations for two cases and applying the given conditions, the value of N can be calculated.

Complete step by step solution:
We know that total current I is given by:
$\text{I =}\dfrac{\text{Total emf}}{\text{Total R}}$
So, for case1, ${{\text{I}}_{1}}=\dfrac{3\text{E}}{\text{R}}$ {As there are three cells connected in series}
We know that
$\text{R}=\dfrac{\text{f l}}{\text{A}}$ ….. (1)
So
${{\text{I}}_{1}}=\dfrac{3\text{E A}}{\text{f l}}$ ….. (2)
The heat produced is given by
$\begin{aligned} & {{\text{H}}_{1}}={{\text{I}}_{1}}^{2}\text{Rt} \\\ & {{\text{H}}_{1}}={{\left( \dfrac{3\text{E A}}{\text{f l}} \right)}^{2}}\left( \dfrac{\text{f l}}{\text{A}} \right)\text{t} \\\ \end{aligned}$
From equation (1) and (2)
$\begin{aligned} & {{\text{H}}_{1}}=\left( \dfrac{9{{\text{E}}^{2}}{{\text{A}}^{2}}}{{{\text{f}}^{2}}\text{ }{{\text{l}}^{2}}} \right)\left( \dfrac{\text{f l}}{\text{A}} \right)\text{t} \\\ & {{\text{H}}_{1}}=\left( \dfrac{9{{\text{E}}^{2}}\text{A}}{{{\text{f}}^{2}}\text{ }} \right)\text{t} \\\ \end{aligned}$
Now heat produced $\left( {{\text{I}}_{1}}^{2}\text{Rt} \right)$ is equal to Q, which is the heat energy and is given by $\text{Q}=\text{mc}\vartriangle \text{T}$
So,
${{\text{I}}_{1}}^{2}\text{Rt}=\text{mc}\vartriangle \text{T}$
$\left( \dfrac{9{{\text{E}}^{2}}\text{At}}{\text{f l}} \right)=\text{mc}\vartriangle \text{T}$ …… (3)
For case 2,
${{\text{I}}_{2}}=\dfrac{\text{NE}}{\text{R}}$ { Here N is to be calculated}
$=\dfrac{\text{N E A}}{\text{f}\left( 2\text{l} \right)}$
So,
${{\text{H}}_{2}}={{\text{I}}_{2}}^{2}\text{Rt}$
$=\left( \dfrac{{{\text{N}}^{2}}{{\text{E}}^{2}}{{\text{A}}^{2}}}{{{\text{f}}^{2}}{{\left( 2\text{ l} \right)}^{2}}} \right)\text{ }\left( \dfrac{\text{f}\left( \text{2 l} \right)}{\text{A}} \right)\text{t}$
${{\text{H}}_{2}}=\left( \dfrac{{{\text{N}}^{2}}{{\text{E}}^{2}}\text{A}}{\text{f }\left( 2\text{ l} \right)} \right)\text{t}$
Now as
${{\text{H}}_{2}}=\text{Q=mc}\vartriangle \text{T}$
$=\left( \dfrac{{{\text{N}}^{2}}{{\text{E}}^{2}}\text{A}}{\text{f}\left( 2\text{ l} \right)} \right)=\left( 2\text{ m} \right)\text{ c}\vartriangle \text{t}$ …… (4) { Here mass is doubled as the length of the wire is double}
On dividing equation (3) by equation (4), we get:
$\left( \dfrac{9{{\text{E}}^{2}}\text{A t}}{\text{f l}} \right)\text{ }\left( \dfrac{\text{f}\left( 2\text{ l} \right)}{{{\text{N}}^{2}}{{\text{E}}^{2}}\text{A t}} \right)=\dfrac{\text{mc}\vartriangle \text{T}}{\left( 2\text{m} \right)\text{c}\vartriangle \text{T}}$
Or
$\dfrac{9\times 2}{{{\text{N}}^{2}}}=\dfrac{1}{2}$
${{\text{N}}^{2}}=36$
$\text{N=6}$

Note: When electric current passes through a conductor, then due to continuous collisions of free electrons with the ions, their energy is repeatedly lost. From ohm’s law, the potential difference between two terminal of a conductor with resistance R when current I is passing through it is V = IR
The magnitude of charge flowing through the conductor in time interval $\vartriangle \text{t}$ is given by:
$\vartriangle \text{q=I}\vartriangle \text{t}$
Electric energy required is calculated as:
$\vartriangle \text{U}$ = Charge $\times$ Potential difference $=\left( \vartriangle \text{q} \right)\times \left( \text{U} \right)$
$\vartriangle \text{U}$ $={{\text{I}}^{2}}\text{R t}$
This electric energy is dissipated in the form of heat
So,
$\begin{aligned} & \text{H}={{\text{I}}^{2}}\text{R t}=\dfrac{{{\text{V}}^{2}}}{\text{R}}\vartriangle \text{t} \\\ & \text{=V I}\vartriangle \text{t } \\\ \end{aligned}$ 2