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  • A wire of length L 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="\Delta" src="https://

A wire of length L and three identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by \DeltaT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross section but of length 2 L. The temperature of the wire is raised by the same amount \DeltaT in the same time t, 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 the wire and let \mathrm{R^{\prime}} be the resistance of the wire. Energy released in t second =\mathrm{\frac{\left(3 V^2\right)}{R \times t }}

\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 }

But \mathrm{Q=m c \Delta T }

\begin{aligned} &\mathrm{ \therefore Q^{\prime}=\frac{\left(N^2 V^2\right)}{2 R} \times t }\\ &\mathrm{ \therefore m c \Delta T=\frac{\left(9 V^2\right)}{R} \times t} \end{aligned}

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

\mathrm{2 m c \Delta T=\frac{\left(N^2 V^2\right)}{2 R} \times t }
Dividing 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 \frac{t}{R}}{N^2 V^2 \frac{t }{2 R}}}\\ &\mathrm{ \therefore \frac{1}{2}=\frac{9 \times 2}{N^2} \Rightarrow N=6}\\ \end{aligned}

Posted by

Suraj Bhandari

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