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  • A wire of length and three identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by <img alt="\mathrm{\Delta T}" src=

A wire of length and three identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by \mathrm{\Delta T} in a time . A number N of similar cells is 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 \mathrm{\Delta T} in the same time , the value of N is

Option: 1

4


Option: 2

6


Option: 3

8


Option: 4

9


Answers (1)

best_answer

Let R be the resistance of rod
Energy released in t seconds = \mathrm{\frac{\left(3 V^2\right)}{R} \times t}

\mathrm{\therefore Q=\frac{\left(9 V^2\right)}{R} \times t }

But \mathrm{Q=m c \Delta T=\frac{\left(9 V^2\right)}{R} \times t \quad \quad \quad (i) }

 Let \mathrm{R^{\prime} } be the resistance of rod


\mathrm{\Rightarrow R^{\prime}=2 R (as~ length ~is ~twice) }

Therefore, energy released in t seconds = \mathrm{\frac{\left(N V^2\right)}{2 R} \times t}


\mathrm{\therefore Q^{\prime}=\frac{\left(N^2 V^2\right)}{2 R} \times t }

Applying \mathrm{Q^{\prime}=m^{\prime} c \Delta T}, we get

\mathrm{2 m c \Delta T=\frac{\left(N^2 V^2\right)}{2 R} \times t }

Diving Eq. (ii) by Eq. (i), we get

\begin{aligned} &\mathrm {\frac{m c \Delta T}{2 m c \Delta T}=\frac{9 V^2 \times t / R}{N^2 V^2 t / 2 R}} \\ & \mathrm{\therefore \frac{1}{2}=\frac{9 \times 2}{N^2} \Rightarrow N^2=18 \times 2 }\\ & \mathrm{\therefore N=6} \end{aligned}

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