Two wires P and Q have equal resistances at \( 20\,^\circ\mathrm{C} \). The temperature coefficients of resistivity of their materials are \( 4.0\times10^{-3}\,{}^\circ\mathrm{C}^{-1} \) and \( 1.0\times10^{-3}\,{}^\circ\mathrm{C}^{-1} \) respectively. Neglecting any change in their dimensions, the ratio \( R_P:R_Q \) of their resistances at \( 70\,^\circ\mathrm{C} \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8 : 7
The size of the wires does not change. So the resistance of each wire changes with temperature in the same way as the resistivity of its material, \( \rho_T=\rho_0[1+\alpha(T-T_0)] \). The two wires start with equal resistances \( R_0 \) at \( 20\,^\circ\mathrm{C} \). The material with the larger temperature coefficient gains more resistance on heating.
The rise in temperature is:
\( \begin{aligned} T-T_0 &= 70\,^\circ\mathrm{C}-20\,^\circ\mathrm{C} \\ &= 50\,^\circ\mathrm{C} \end{aligned} \)
For wire P:
\( \begin{aligned} R_P &= R_0\,[1+4.0\times10^{-3}\times50] \\ &= 1.20\,R_0 \end{aligned} \)
For wire Q:
\( \begin{aligned} R_Q &= R_0\,[1+1.0\times10^{-3}\times50] \\ &= 1.05\,R_0 \end{aligned} \)
The ratio of the two resistances is:
\( \begin{aligned} \frac{R_P}{R_Q} &= \frac{1.20}{1.05} \\ &= \frac{8}{7} \end{aligned} \)
Therefore, \( R_P:R_Q=8:7 \) at \( 70\,^\circ\mathrm{C} \).