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A $100 \mathrm{~W}$ tungsten light bulb has a resistance of $250 \Omega$ when it as turned ON and $25 \Omega$ when turned OFF. The ambient room temperature is $25^{\circ} \mathrm{C}$. Find the temperature of the filament when the bulb is turned ON.
$\left(\right.$ Let $\left.\alpha_{\text {tungsten }}=4.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}\right)$
Options:
$\left(\right.$ Let $\left.\alpha_{\text {tungsten }}=4.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}\right)$
Solution:
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Verified Answer
The correct answer is:
$2025^{\circ} \mathrm{C}$
If $t_2$ and $t_1$ be the temperature of filament,
when light bulb is turned ON and turned OFF, respectively
then,
$\begin{aligned} R_{t_2} & =250 \Omega \\ R_{t_1} & =25 \Omega \\ \alpha_{\text {tungsten }} & =4.5 \times 10^{-3} / \mathrm{C}\end{aligned}$
We know that, $\alpha=\frac{R_{t_2}-R_{t_1}}{R_{t_1}\left(t_2-t_1\right)}$
Putting the given values, we get
$4.5 \times 10^{-3}=\frac{250-25}{25\left(t_2-25\right)}$
$\begin{aligned} 4.5 \times 10^{-3} & =\frac{9}{t_2-25} \\ 4.5 \times 10^{-3} t_2-112.5 \times 10^{-3} & =9 \\ 4.5 t_2-112.5 & =9 \times 10^3 \\ 4.5 t_2 & =9000+112.5 \\ 4.5 t_2 & =9112.5 \\ t_2 & =\frac{9112.5}{4.5}=2025^{\circ} \mathrm{C}\end{aligned}$
Hence, the temperature of the filament when the bulb is turned on $2025^{\circ} \mathrm{C}$.
when light bulb is turned ON and turned OFF, respectively
then,
$\begin{aligned} R_{t_2} & =250 \Omega \\ R_{t_1} & =25 \Omega \\ \alpha_{\text {tungsten }} & =4.5 \times 10^{-3} / \mathrm{C}\end{aligned}$
We know that, $\alpha=\frac{R_{t_2}-R_{t_1}}{R_{t_1}\left(t_2-t_1\right)}$
Putting the given values, we get
$4.5 \times 10^{-3}=\frac{250-25}{25\left(t_2-25\right)}$
$\begin{aligned} 4.5 \times 10^{-3} & =\frac{9}{t_2-25} \\ 4.5 \times 10^{-3} t_2-112.5 \times 10^{-3} & =9 \\ 4.5 t_2-112.5 & =9 \times 10^3 \\ 4.5 t_2 & =9000+112.5 \\ 4.5 t_2 & =9112.5 \\ t_2 & =\frac{9112.5}{4.5}=2025^{\circ} \mathrm{C}\end{aligned}$
Hence, the temperature of the filament when the bulb is turned on $2025^{\circ} \mathrm{C}$.
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