Resistance of a tungsten wire at \(150^{\circ} \mathrm{C}\) is resistance is \(0.0045^{\circ}…
Resistance of a tungsten wire at \(150^{\circ} \mathrm{C}\) is resistance is \(0.0045^{\circ} \mathrm{C}^{-1}\). The resistance of this wire at \(500^{\circ} \mathrm{C}\) is
\(180 \Omega\)
\(225 \Omega\)
\(258 \Omega\)
\(317 \Omega\)
Solution
Resistance of tungsten wire at $150^{\circ} \mathrm{C}$ is given as
$\begin{aligned}
& R_{150}=133 \Omega \\
& \Rightarrow \quad R_0(1+\alpha t)=133 \quad\left[\because R_t=R_0(1+\alpha t)\right] \\
& \Rightarrow \quad R_0(1+150 \alpha)=133 \quad \ldots (i) \\
\end{aligned}$
Similarly, resistance of wire at 500^{\circ} \mathrm{C} is given as
$\begin{aligned}
& R_{500}=R_0(1+500 \alpha) \\
& R_0(1+500 \alpha)=R_{500} \quad \ldots (ii)
\end{aligned}$
From Eqs. (i) and (ii), we get
$\begin{aligned}
& \frac{R_0(1+150 \alpha)}{R_0(1+500 \alpha)}=\frac{133}{R_{500}} \\
\Rightarrow \quad & R_{500}=\frac{133(1+500 \alpha)}{1+150 \alpha}=\frac{133(1+500 \times 0.0045)}{1+150 \times 0.0045} \quad {\left[\text {given, } \alpha=0.0045^{\circ} \mathrm{C}^{-1}\right] } \\
& =\frac{43225}{1.675}=258.06 \Omega \simeq 258 \Omega
\end{aligned}$