1. Tardigrade
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  3. Physics
  4. The resistance of a wire is 5 Ω at 50 ° C and 6 Ω at 100 ° C. The resistance of the wire at 0 ° C will be

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Q. The resistance of a wire is $5 \Omega$ at $50{ }^{\circ} C$ and $6 \quad \Omega$ at $100{ }^{\circ} C$. The resistance of the wire at $0{ }^{\circ} C$ will be

NTA AbhyasNTA Abhyas 2022

Solution:

$ R _{ t }= R _{0}(1+\alpha t ) $
Where $R _{ t }=$ resistance of a wire at $t \quad{ }^{\circ} C , R _{0}=$ resistance of a wire at $0{ }^{\circ} C$, $\alpha=$ temperature coefficient of resistance.
$ \therefore R _{50}= R _{0}[1+\alpha 50] $
and $R _{100}= R _{0}[1+\alpha 100]$
or $R _{50}- R _{0}= R _{0} \alpha(50)$
$ R _{100}- R _{0}= R _{0} \alpha(100) $
Dividing (i) by (ii), we get
$ \frac{5-R_{0}}{6-R_{0}}=\frac{1}{2} $
or $\quad 10-2 R_{0}=6-R_{0}$
or $R_{0}=4 \Omega$