The same mass of copper is drawn into two wires 1mm and 2mm thick. Two wires are connected in series and current is passed through them. Heat produced in the wires is in the ratio

2:1

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1:16

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4:1

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16:1

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Solution

The correct option is D 16:1
same mass $\Rightarrow$ same volume

$l_{1} A_{1} = l_{2} A_{2}$

$R_{1} = ρ \frac{l_{1}}{A_{1}}$

$R_{2} = ρ \frac{l_{2}}{A_{2}}$
ration of heat produced = $\frac{I^{2} R_{1}}{I^{2} R_{2}} = \frac{l_{1} A_{2}}{A_{1} l_{2}} = (\frac{A_{2}}{A_{1}})^{2} = (\frac{r_{2}^{2}}{r_{1}^{2}})^{2} = (\frac{r_{2}}{r_{1}})^{4} = (2)^{4} = 16$

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The same mass of copper is drawn into two wires 1mm and 2mm thick. Two wires are connected in series and current is passed through them. Heat produced in the wires is in the ratio

The correct option is D 16:1same mass ⇒ same volumel1A1=l2A2R1=ρl1A1R2=ρl2A2ration of heat produced = I2R1I2R2=l1A2A1l2=(A2A1)2=(r22r21)2=(r2r1)4=(2)4=16