Question

A wire of length $L$ and $3$ identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by $\Delta T$ in time $t$. $N$ number of similar cells are now connected in series with a wire of the same material and cross-section but of length $2L$. The temperature of the wire is raised by the same amount $\Delta T$ in the same time $t$. The value of $N$ is:

• A. $4$
• B. $6$
• C. $8$
• D. $9$

Answer 1

The correct option is B $6$

Let $A$ be the cross-sectional area and $\rho$ be the resistivity of the material.
$\therefore R=\frac{\rho L}{A}$

And resistance of wire of $2L$ length is

$R^{\prime }=\frac{\rho \left(2L\right)}{A}=2R$

Let emf of each cell be $E$.

When $3$ identical cells are connected in series.

So net emf, $E_1=3E$

Since, in time $t,\Delta T$ temperature raised due to heat generation $(H=V^2/R)$.

$\frac{(3E{)}^2}{R}=ms\Delta T......\left(1\right)$

And when $N$ identical cells are connected in series.

Net emf $E_2=NE$

$\therefore \frac{(NE{)}^2}{2R}=\left(2m\right)s\Delta T......\left(2\right)$

Dividing eq. (1) to (2), we get

$\frac{N^2}{9}=\frac{4}{1}\Rightarrow N^2=36$

$\therefore N=6$

Hence, option $\left(b\right)$ is correct.