Question

Temperature coefficient of resistance of platinum is $4 \times 10^{-3} / K$ at $20^{\circ} C$. Temperature at which increase in resistance of platinum is $10 \%$ its value at $20^{\circ} C$ is

• A. $25^{\circ} C$
• B. $70^{\circ} C$
• C. $45^{\circ} C$
• D. $100^{\circ} C$

Answer 1

Answer: (C)

Here, $\alpha = 4 \times 10^{-3} \, K^{-1}$ at $T_0 = 20^{\circ} C$
Also, $R_t = R_0(1 + \alpha \Delta T)$
Here, $R_t = R_0 + 10\%$ of $R_0 = \frac{110}{100} R_0 = \frac{11}{10} R_0$
$\therefore \, \, \frac{11}{10} R_0 = R_0 (1 + 4 \times 10^{-3} \Delta T )$
or $\frac{11}{10} = 1 + 4 \times10^{-3} \Delta T$
or $4 \times10^{-3} \Delta T = \frac{1}{10}$
or $\Delta T =\frac{1}{4 \times10^{-2}} = 25^{\circ}C$
or$\Delta T = T - T_{0} + 25^{\circ}C$
$\therefore \, T = T_{0 } + 25 = 20 + 25 = 45^{\circ}C$