Q147 / 180 · KCET 2015 · Physics

The resistance of the bulb filament is \( 100 \Omega \) at a temperature of \( 100^{\circ} \mathrm{C} \). If its temperature co-
efficient of resistance be \( 0.005 \) per \( { }^{\circ} \mathrm{C} \), its resistance will become \( 200 \Omega \) at a temperature

  • Correct answer
    A
    \( 300^{\circ} \mathrm{C} \)
  • B
    \( 400^{\circ} \mathrm{C} \)
  • C
    \( 500^{\circ} \mathrm{C} \)
  • D
    \( 200^{\circ} \mathrm{C} \)

Solution

At $T_{1}=100^{\circ} \mathrm{C}$, resistance, $R_{1}=100 \Omega ;$ temperature coefficient, $\alpha=0.005$ per ${ }^{\circ} \mathrm{C}$ Let resistance $R_{2}=200 \Omega$ at temperature $T_{2}$, then $R_{2}=R_{1}\left(1+\alpha\left(T_{2}-T_{1}\right)\right) \Rightarrow R_{2}=R_{1}+\alpha R_{1}\left(T_{2}-T_{1}\right)$ $\Rightarrow R_{2}-R_{1}=\alpha R_{1}\left(T_{2}-T_{1}\right) \Rightarrow \frac{\left(R_{2}-R_{1}\right)}{\alpha R_{1}}=T_{2}-T_{1}$ $\Rightarrow T_{2}=T_{1}+\frac{\left(R_{2}-R_{1}\right)}{\alpha R_{1}}$ Substituting the values, we get $T_{2}=100+\frac{(200-100)}{0.005 \times 100}=300^{\circ} \mathrm{C}$

Hence, the correct answer is option: 'A'.

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