Question

Ten identical cells connected in series are needed to heat a wire of length one meter and radius '$r$' by $10^{\circ }C$ in time '$t$'. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time '$t$' ?

• A. 20
• B. 30
• C. 40
• D. 10

Answer 1

Answer: (A)

Let $\rho$ be resistivity of the material of the wire and $r$ be radius of the wire.
Therefore, resistance of $1m$ wire is
$R=\frac{\rho (1)}{\pi r^2}=\frac{\rho }{\pi r^2}$
$\left(\because R=\frac{\rho l}{A}\right)$
Let $\epsilon$ be emf of each cell.
In first case,

$10$ cells each of emf $\epsilon$ are connected in series to heat the wire of length $1m$ by $\Delta T\left(=10^{\circ }C\right)$ in time $t$.
$\therefore \frac{(10\epsilon {)}^2}{R}t=ms\Delta T\dots \text{ (i) }$
In second case,
Resistance of same wire of length $2m$ is
$R^{\prime }=\frac{\rho (2)}{\pi r^2}=\frac{2\rho }{\pi r^2}=2R$

Let $n$ cells each of emf $\epsilon$ are connected in series to heat the same wire of length $2m$, by the same temperature $\Delta T\left(=10^{\circ }C\right)$ in the same time $t$.
$\therefore \frac{(n\epsilon {)}^{2}t}{2R}=(2m)s\Delta T\dots \text{ (ii) }$
Divide (ii) by (i), we get
$\frac{n^2}{200}=2$
$\Rightarrow n^2=400$
$\therefore n=20$