A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by $∆$ T in a 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 2 I- The temperature of the wire is raised by the same amount $∆$ T in the same time t. The value of N is

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answer is B.

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Detailed Solution

We know that heat developed ${=i}^{2}$ R t. Hence
${(\frac{3 E}{R_{1}})}^{2} R_{1} t = m_{1} sdt$
and ${(\frac{NE}{R_{2}})}^{2} R_{2} t = m_{2} sdt$
$∴ \frac{N^{2}}{9} = \frac{m_{2}}{m_{1}} \times \frac{R_{2}}{R_{1}}$
or $N^{2} = 9 (\frac{m_{2}}{m_{1}}) (\frac{R_{2}}{R_{1}})$
$N = 3 \sqrt{(\frac{m_{2}}{m_{1}}) (\frac{R_{2}}{R_{1}})} = 3 \sqrt{(\frac{2 m_{1}}{m_{1}}) (\frac{2 R_{1}}{R_{1}})} = 6$