A 100 W tungsten light bulb has a resistance of `250Omega` when it as turned ON and `25Omega` when turned OFF. The ambient room temperture is `25^(@)C`. Find the temperature of the filament when the bulb is turned ON, ( Let `alpha_("tungsten")=4.5xx10^(-3)//^(@)C`)

Options

A. `2600^(@)C`
B. `2025^(@)C`
C. `2500^(@)C`
D. `2625^(@)C`

Answer

B

Explanation

If `t_(2)andt_(1)` be the temperature of filament, when light bulb is turned ON and turned OFF, respectively
then, `R_(t_(2))=250Omega`
`R_(t_(1))=25Omega`
`alpha_("tungsten")=4.5xx10^(-3)l^(@)C`
We know that, `alpha=(R_(t_(2))-R_(t_(1)))/(R_(t_(1))(t_(2)-t_(1)))`
Putting the given values, we get
`4.5xx10^(-3)=(250-25)/(25(t_(2)-25))`
`4.5xx10^(-3)=(9)/(t_(2)-25)`
`4.5xx10^(-3)t_(2)-112.5xx10^(-3)=9`
`4.5t_(2)-112.5=9xx10^(3)`
`4.5t_(2)=9000+112.5`
`4.5t_(2)=9112.5`
`t_(2)=(9112.5)/(4.5)=2025^(@)C`
Hence, the temperature of the filament when the bulb is turned on `2025^(@)C`.
Hence, the temperature of the filament when the bulb is turned on `2025^(@)C`