© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Physical Quantities
We are given:
• A wire of length 1 m and radius $r$ requiring 10 cells in series to be heated by $10^\circ C$ in time $t.$
• The same wire of length 2 m and radius $r,$ with the same temperature rise ($10^\circ C$) and same time $t,$ but an unknown number of cells $n.$
Step 2: Write the Resistance of the Wire
The resistance $R$ of a uniform wire of length $l,$ cross-sectional area $A,$ and resistivity $\rho$ is given by
$$
R = \frac{\rho\,l}{A} = \frac{\rho\,l}{\pi r^2}.
$$
For the 1 m wire:
$$
R = \frac{\rho \times 1}{\pi r^2} = \frac{\rho}{\pi r^2}.
$$
Step 3: Set Up the Heating (Energy) Equation for the 1 m Wire
If each cell has an emf $\varepsilon,$ then 10 cells in series provide a total emf of $10\varepsilon.$
The power delivered when a potential $V$ (here $10\varepsilon$) is applied across a resistor $R$ is
$$
P = \frac{(10\varepsilon)^2}{R}.
$$
Since heating is the energy supplied in time $t,$ we write
$$
\text{Energy} = P \times t = \frac{(10\varepsilon)^2}{R} \, t.
$$
This energy goes into heating the wire by $\Delta T = 10^\circ C.$ Let $m$ be the mass of the wire and $s$ its specific heat. Then,
$$
\frac{(10\varepsilon)^2}{R} \, t = m\,s\,\Delta T.
$$
(Equation i)
Step 4: Resistance for the 2 m Wire
When the length is doubled to $2$ m (keeping the same radius $r$), the new resistance $R'$ becomes
$$
R' = \frac{\rho \times 2}{\pi r^2} = 2R.
$$
Step 5: Heating Equation for the 2 m Wire
If $n$ cells are connected in series, the total emf is $n\varepsilon.$ The power is
$$
P' = \frac{(n\varepsilon)^2}{R'}.
$$
Over the same time $t,$ the energy provided is
$$
\frac{(n\varepsilon)^2}{R'} \, t = \frac{(n\varepsilon)^2}{2R} \, t.
$$
The wire is now twice as long, so its mass doubles (assuming the same material and cross-sectional area). The energy needed for the same temperature rise $\Delta T$ is
$$
2\,m\,s\,\Delta T.
$$
Hence,
$$
\frac{(n\varepsilon)^2}{2R} \, t = 2\,m\,s\,\Delta T.
$$
(Equation ii)
Step 6: Form the Ratio of the Two Heating Equations
Dividing (ii) by (i):
$$
\frac{\frac{(n\varepsilon)^2}{2R} \, t}{\frac{(10\varepsilon)^2}{R} \, t} = \frac{2\,m\,s\,\Delta T}{m\,s\,\Delta T}.
$$
Simplify:
$$
\frac{(n\varepsilon)^2}{2R} \,\Big/\, \frac{(10\varepsilon)^2}{R} = 2.
$$
$$
\frac{(n\varepsilon)^2}{2R} \times \frac{R}{(10\varepsilon)^2} = 2.
$$
$$
\frac{n^2 \varepsilon^2}{(10)^2 \varepsilon^2} \times \frac{1}{2} = 2.
$$
$$
\frac{n^2}{100} \times \frac{1}{2} = 2.
$$
$$
\frac{n^2}{200} = 2.
$$
$$
n^2 = 400.
$$
Hence,
$$
n = 20.
$$
Final Answer
The required number of cells is 20.
