question_answer

A flash light tamp is marked 3.5 V and 0.28 A. The filament temperature is 425°C. The filament resistance at 0°C is 4°. Then, [he temperature coefficient of resistance of the material of die filament is

A) \[8.5\times {{10}^{-3}}/K\]                         

B)  \[3.5\times {{10}^{-3}}/K\]

C) \[0.5\times {{10}^{-3}}/K\]                         

D)  \[5\times {{10}^{-3}}/K\]

Correct Answer: D

Solution :

 Here, \[t=425{{\,}^{o}}C,{{R}_{0}}=4\Omega \] By ohms law \[V=I{{R}_{t}}\] \[\therefore \]  \[3.5=0.28{{R}_{t}}\]                 \[\frac{350}{28}={{R}_{t}}\]                 \[{{R}_{t}}=12.5\Omega \] Resistance at temperature \[t{{\,}^{o}}C\]is, \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\]where\[\alpha \]is temperature coefficient of resistance. \[\therefore \]  \[12.5=4(1+\alpha \times 425)\] or            \[\frac{12.5}{4}-1=\alpha \times 425\] or            \[\frac{8.5}{4}=\alpha \times 425\] \[\alpha =\frac{8.5}{4\times 425}\] \[=0.5\times {{10}^{-2}}\] \[=5\times {{10}^{-3}}/K\]