QJEEM24Phc12c: JEE Main 2024 Physics - Class 12 JEEM24Phc12c Solution: Resistance of a wire at and is found to be and respectively.…

To determine the temperature $t$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:

$R_t = R_0 (1 + \alpha t)$

where:

• $R_t$ is the resistance at temperature $t$
• $R_0$ is the resistance at the reference temperature (usually $0^{\circ} \mathrm{C}$ )
• $\alpha$ is the temperature coefficient of resistance
• $t$ is the temperature

We are given the following resistances:

• Resistance at $0^{\circ} \mathrm{C}$ : $R_0 = 10 \Omega$
• Resistance at $100^{\circ} \mathrm{C}$ : $R_{100} = 10.2 \Omega$
• Resistance at $t^{\circ} \mathrm{C}$ : $R_t = 10.95 \Omega$

First, we need to find the temperature coefficient of resistance $\alpha$ . Using the resistance at $100^{\circ} \mathrm{C}$ :

$10.2 = 10 (1 + \alpha \cdot 100)$

Solving for $\alpha$ :

$\frac{10.2}{10} = 1 + 100\alpha \implies 1.02 = 1 + 100\alpha \implies 100\alpha = 0.02 \implies \alpha = \frac{0.02}{100} = 0.0002$

Now, we can find the temperature $t$ using the resistance at $t^{\circ} \mathrm{C}$ :

$10.95 = 10 (1 + 0.0002 \cdot t)$

Solving for $t$ :

$\frac{10.95}{10} = 1 + 0.0002 t \implies 1.095 = 1 + 0.0002 t \implies 0.0002 t = 0.095 \implies t = \frac{0.095}{0.0002} = 475$

The temperature $t$ in Celsius is $475^{\circ} \mathrm{C}$ . To convert this to the Kelvin scale:

$T_{K} = t_{C} + 273.15 = 475 + 273.15 = 748.15 \, \mathrm{K}$

So, the temperature $t$ in Kelvin scale is approximately $748.15 \, \mathrm{K}$ .