Numerical Problems Based on Temperature Variation of Resistance for Class 12 Physics

Initial resistance $R_0 = 10 \Omega$ at $t_1 = 0^{\circ}\text{C}$.
Final resistance $R_t = 20 \Omega$ at $t_2 = 273^{\circ}\text{C}$.

The variation of resistance with temperature is given by:

Substitute the given values:

Initial temperature $t_1 = 20^{\circ}\text{C}$. Let the final temperature where resistances are equal be $t^{\circ}\text{C}$. The change in temperature is $\Delta t = t – 20$.

For iron: $R_{Fe} = 3.9(1 + 5.0 \times 10^{-3} \Delta t)$
For copper: $R_{Cu} = 4.1(1 + 4.0 \times 10^{-3} \Delta t)$

Equating the two resistances ($R_{Fe} = R_{Cu}$):

The final temperature $t$ is:

Initial resistance $R_0 = 1.25 \Omega$ at $0^{\circ}\text{C}$.
Final resistance $R_t = 2 \times 1.25 = 2.50 \Omega$.
Temperature coefficient $\alpha = 0.00375 ^{\circ}\text{C}^{-1}$.

Using the relation $R_t = R_0(1 + \alpha \Delta t)$:

Since the initial temperature was $0^{\circ}\text{C}$, the final temperature is $267^{\circ}\text{C}$.

Yes, the temperature will be the same for all silver conductors of all shapes because the fractional change in resistance ($\frac{\Delta R}{R}$) depends only on the temperature coefficient of the material and the temperature change, not on physical dimensions.

Resistance at ice point $R_0 = 3.00 \Omega$.
Resistance at steam point $R_{100} = 3.75 \Omega$.
Resistance at unknown temperature $R_t = 3.15 \Omega$.

The unknown temperature $t$ on the platinum scale is given by the formula:

Substitute the given values:

Temperature coefficient $\alpha = 0.00125 ^{\circ}\text{C}^{-1}$.
Initial temperature $T_1 = 300\text{ K} = 300 – 273 = 27^{\circ}\text{C}$. Resistance $R_1 = 1 \Omega$.
Final resistance $R_2 = 2 \Omega$. Let final temperature be $t_2 ^{\circ}\text{C}$.

Expressing resistance at temperatures $27^{\circ}\text{C}$ and $t_2$ in terms of $R_0$ (resistance at $0^{\circ}\text{C}$):

Dividing the second equation by the first:

Converting back to Kelvin:

Given $\alpha = 0.004 ^{\circ}\text{C}^{-1}$, $l = 5\text{ m}$, diameter $d = 0.2\text{ mm} \implies r = 0.1\text{ mm} = 10^{-4}\text{ m}$.
Resistivity at $0^{\circ}\text{C}$, $\rho_0 = 1.7 \times 10^{-8} \Omega\text{ m}$.

1. Calculate resistivity at $100^{\circ}\text{C}$ ($\rho_{100}$):

2. Calculate the resistance at $100^{\circ}\text{C}$ ($R$):

Area $A = \pi r^2 = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8}\text{ m}^2$.

Rounding to proper significant figures: