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The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. The resistance of the wire at 0°C will be

Option 1)
$3\: \: ohm$

Option 2)
$2\: \: ohm$

Option 3)
$1\: \: ohm$

Option 4)
$4\: \: ohm$

Answers (2)

best_answer

As we learnt in

Temperature co-efficient of Resistivity -

$\alpha=\frac{R_{T}-R_{o}}{R_{o}(T-T_{o})}$

- wherein

Where the value of $\alpha$ is different at different temperatures

Given : $R_{50}=5\Omega ,R_{100}=6\Omega$

$R_{t}=R_{0}\left ( 1+\alpha t \right )$

Where $R_{t}$ = resistance of a wire at $t^{\circ}C$

$R_{0}$ = resistance of a wire at $0^{\circ}C$ , $\alpha =$ Temperature coefficient of resistance.

$\therefore R_{50}= R_{0}\left [ 1+\alpha 50 \right ]$

$R_{100}= R_{0}\left [ 1+\alpha 100 \right ]$

$or\: \: \: \: R_{50}- R_{0}= R_{0}\alpha \left ( 50 \right )\cdots \cdots \cdots \cdots \left ( i \right )$

$or\: \: \: \: R_{100}- R_{0}= R_{0}\alpha \left ( 100 \right )\cdots \cdots \cdots \cdots \left (ii \right )$

Divide (i) by (ii), we get

$\frac{5-R_{0}}{6-R_{0}}= \frac{1}{2}$

$or 10-2R_{0}= 6-R_{0}$

$or\: \: \: \: R_{0}=4\Omega$

Option 1)

$3\: \: ohm$

This option is incorrect.

Option 2)

$2\: \: ohm$

This option is incorrect.

Option 3)

$1\: \: ohm$

This option is incorrect.

Option 4)

$4\: \: ohm$

This option is correct

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