The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300K its resistance is 1Ω. The resistance of wire will be 2Ω at

The correct option is 1127 K.

Step-by-Step Explanation:

The variation of resistance with temperature is described by the relation:

$R\left(t\right)=R_0$

where:
- $R_0$ is the resistance of the wire at the reference temperature $0\text{°C}$ (or $273\text{K}$).
- $\alpha$ is the temperature coefficient of resistance, given as $0.00125{\text{°C}}^{-1}$.
- $t$ is the temperature in degrees Celsius (°C).

First, we convert the given initial temperature from Kelvin to Celsius:

$t_1=300-273=27\text{°C}$

At $t_1=27\text{°C}$, the resistance of the wire is $R_1=1\text{Ω}$. We write this relation as:

$1=R_0(1+\alpha t_1)$ ��— (Equation 1)

Let the target resistance be $R_2=2\text{Ω}$ at a temperature $t_2$ in Celsius. We write the relation for this state as:

$2=R_0(1+\alpha t_2)$ —— (Equation 2)

Dividing Equation 2 by Equation 1 to eliminate the reference resistance $R_0$:

$\frac{2}{1}=\frac{1+\alpha t_2}{1+\alpha t_1}$

Cross-multiplying, we obtain:

$2(1+\alpha t_1)=1+\alpha t_2$

$2+2\alpha t_1=1+\alpha t_2$

Rearranging the equation to solve for $t_2$:

$\alpha t_2=1+2\alpha t_1$

$t_2=\frac{1}{\alpha }+2t_1$

Substitute the value of $\alpha =0.00125=\frac{1}{800}$ and $t_1=27\text{°C}$ into the equation:

$t_2=800+2\left(27\right)$

$t_2=800+54=854\text{°C}$

Now, convert the final temperature from Celsius back to Kelvin:

$T_2=t_2+273$

$T_2=854+273=1127\text{K}$

Therefore, the resistance of the wire will be 2Ω at a temperature of 1127 K.