JEE Main 2024 — Current Electricity Question with Solution

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 $$\alpha$$:

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

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 $$t$$:

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

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 $$748.15\mathrm{K}$$.