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

16 : 1

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8 : 32

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8 : 2

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4 : 2

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Solution

The correct option is A $16 : 1$
Same mass $\Rightarrow$ Same volume

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

$R_{1} = ρ \frac{l_{1}}{A_{1}}$

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

So, the heat produced is in the ratio of $16 : 1$

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

The correct option is A 16:1Same mass ⇒ Same volumel1A1=l2A2R1=ρl1A1R2=ρl2A2Ratio of heat produced = I2R1I2R2=l1A2A1l2=(A2A1)2=(r22r21)2=(r2r1)4=(12)4=161So, the heat produced is in the ratio of 16:1

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