Question

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

Answer 1

Hint: We know that the relation between resistance and length can be given by the formula, $R=\dfrac{\rho l}{A}$, using this formula we will find the relation between wire 1 and 2, then we will find the heat generated or dissipated using the formula, $H={{I}^{2}}Rt$. Then we will replace the value of current (I) with voltage difference (V) by using the formula, $I=\dfrac{V}{R}$. And then by equating the expressions we will find the value of N.

Formula used: $R=\dfrac{\rho l}{A}$, $H={{I}^{2}}Rt$, $I=\dfrac{V}{R}$

Complete step by step answer:
In question it is given that a wire of length L and three identical cells of negligible internal resistance are connected in series. Due to the current, the temperature of the wire is raised by $\Delta T$ in time t. A number N of similar cells is now connected in series with a wire of same material and cross section but of length $2L$. The temperature of the wire is raised by the same amount $\Delta T$ at the same time. So, now, from this we have two different wires of different length but having the same cross-sectional area and same material.
Now, we know that resistance of wire can be given by the formula,
$R=\dfrac{\rho l}{A}$ ………………..(i)
Where, R is resistance of wire, l is length, A is cross sectional area and $\rho $ is resistivity of the wire.
For wire 1, ${{R}_{1}}=\dfrac{\rho {{l}_{1}}}{{{A}_{1}}}$
Where, ${{l}_{1}}=L$, remaining all remains the same. So,
${{R}_{1}}=\dfrac{\rho L}{A}$
For wire 2, ${{R}_{2}}=\dfrac{\rho {{l}_{2}}}{{{A}_{2}}}$
Where, ${{l}_{2}}=2L$ rest all remains the same.
So, ${{R}_{2}}=\dfrac{\rho 2L}{A}$
${{R}_{2}}=\dfrac{\rho 2L}{A}\Rightarrow {{R}_{2}}=2\left( \dfrac{\rho L}{A} \right)=2{{R}_{1}}$ ……………………(ii)
Now, the heat generated due to resistance of wire can be give by the formula,
$H={{I}^{2}}Rt$ ……………………(iii)
where, H is heat generated, I is current passing through the wire, R is resistance of the wire and t is time.
Heat generated can also be given by the formula,
$H=mc\Delta T$ ……………………..(iv)
Where, m is mass, c is coefficient of heat energy and T is temperature.
For, wire 1, 3 cells are connected in series so the voltage difference across the wire will be $3V$and the relation between voltage, current and resistance can be given as,
$I=\dfrac{V}{R}$
Now, substituting value of I in expression (ii) for wire 1, we will get,
$H={{I}^{2}}Rt$
$H={{\left( \dfrac{3V}{{{R}_{1}}} \right)}^{2}}{{R}_{1}}t=mc\Delta T$………..(v)
Now, for wire 2, it is said that n cells are connected in series so, voltage difference will be $NV$, on substituting the value in expression (iii) we will get,
$H={{\left( \dfrac{NV}{{{R}_{2}}} \right)}^{2}}{{R}_{2}}t=mc\Delta T$……………(vi)
On, dividing the expressions (v) and (vi) we will get,
$\dfrac{{{\left( \dfrac{3V}{{{R}_{1}}} \right)}^{2}}{{R}_{1}}t}{{{\left( \dfrac{NV}{{{R}_{2}}} \right)}^{2}}{{R}_{2}}t}=\dfrac{mc\Delta T}{mc\Delta T}$
Now, on substituting the value of resistance of wire 2 from expression (ii) we will get,
$\Rightarrow \dfrac{9{{V}^{2}}}{{{R}_{1}}}=\dfrac{{{N}^{2}}{{V}^{2}}}{4{{R}_{1}}}$
$\Rightarrow {{N}^{2}}=9\times 4\Rightarrow N=3\times 2=6$
Hence, the value of N is 6.
Thus, option (b) is the correct answer.

Note: Students must know the concept of series and parallel i.e. in series voltage is added and current remains same while in parallel voltage across each wire remains same and current is added. Here we have added the voltage of cells because they are connected in series, now if the cells were in parallel the voltage across each cell would be the same or constant. So, students must know these concepts of series and parallel before solving it.