QJEEM24Phd7ce: JEE Main 2024 Physics - Class 11 JEEM24Phd7ce Solution: The resistances of the platinum wire of a platinum resistance…

The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:

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

where:

• $R_t$ is the resistance at temperature $t$ ,


• $R_0$ is the resistance at 0°C (ice point),


• $\alpha$ is the temperature coefficient of resistance, and


• $t$ is the temperature in degrees Celsius.

In this question, we are given:

• Resistance at the ice point ( $R_{0} = 8 \Omega$ ),


• Resistance at the steam point ( $R_{100} = 10 \Omega$ ).

To find the temperature coefficient of resistance ( $\alpha$ ), we use the resistance values at the ice and steam points:

$\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C$

Now, to find the resistance $R_t$ at $400 ^{\circ}C$ , we substitute $\alpha$ , $R_0$ , and $t = 400$ into the formula:

$R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega$

Hence, the resistance of the platinum wire at $400^{\circ}C$ is $16 \Omega$ . The correct option is:

Option B: 16 $\Omega$