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 $\frac{11}{10}=1+4\times 10^{-3}\Delta T$
or $4\times 10^{-3}\Delta T=\frac{1}{10}$
or $\Delta T=\frac{1}{4\times 10^{-2}}=25^{\circ }C$
or$\Delta T=T-T_0+25^{\circ }C$
$\therefore T=T_0+25=20+25=45^{\circ }C$