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Question: The temperature coefficient resistivity of a material is \(0.0004/K\). When the temperature of the m...

The temperature coefficient resistivity of a material is 0.0004/K0.0004/K. When the temperature of the material is increased by 50∘C{50^\circ }C , its resistivity increases by 2×10−82 \times {10^{ - 8}}. The initial resistivity of the material of the resistance is:
A) 50×10−850 \times {10^{^{ - 8}}}
B) 90×10−890 \times {10^{ - 8}}
C) 100×10−8100 \times {10^{ - 8}}
D) 200×10−8200 \times {10^{ - 8}}

Explanation

Solution

In the given question, we need to find the initial resistivity of the material given the rise in resistivity due to the increase in temperature, this can be done using the relation, as we know, and the initial resistivity is directly proportional to the rise in temperature.

Formula used:
ρt=ρ0[1+αΔT]{\rho _t} = {\rho _0}[1 + \alpha \Delta T]
where,
ρt={\rho _t} = Resistivity at standard temperature
ρ0={\rho _0} = Initial resistivity
α=\alpha = Coefficient of resistivity
ΔT=\Delta T = Rise in temperature

Complete step by step solution:
Resistivity can be defined as the resistance of a conductor having unit length and unit cross-sectional area. This is a characteristic property of the material, and thus it is directly proportional to temperature.
As temperature increases, there is a rise in resistivity, similarly, with a decrease in temperature, the resistivity of the conductor also decreases.
Thus, we can say that the resistivity ff a conductor depends on the material and temperature but is independent of the shape and size of the conductor.
We can say, ρ∝ΔT\rho \propto \Delta T
where,
ρ=\rho = Resistivity of the material
ΔT=\Delta T = Rise in temperature
We know the formula that states the relation between resistivity and rise in temperature goes as:
ρt=ρ0[1+αΔT]{\rho _t} = {\rho _0}[1 + \alpha \Delta T]
where,
ρt={\rho _t} = Resistivity at standard temperature
ρ0={\rho _0}= Initial resistivity
α=\alpha = Coefficient of resistivity
ΔT=\Delta T = Rise in temperature
On rearranging the equation, we obtain:
⇒ρt−ρ0=ρ0αΔT\Rightarrow {\rho _t} - {\rho _0} = {\rho _0}\alpha \Delta T
We know:
⇒ρt−ρ0=\Rightarrow {\rho _t} - {\rho _0}= Rise in resistivity
In the question, we have the value for the change in resistivity,
Putting the values, in the equation:
⇒2×10−8=ρ0×0.0004×50\Rightarrow 2 \times {10^{ - 8}} = {\rho _0} \times 0.0004 \times 50
⇒ρ0=2×10−80.0004×50\Rightarrow {\rho _0} = \dfrac{{2 \times {{10}^{ - 8}}}}{{0.0004 \times 50}}
Thus, on solving the equation we obtain:
⇒ρ0=100×10−8ohm−m\Rightarrow {\rho _0} = 100 \times {10^{ - 8}}ohm - m
This is the required solution.

Thus, option (C ) is correct.

Note: Although, with an increase in temperature resistivity increases however resistance decreases. This happens because, as the temperature rises, the band gap between conduction and valence band decreases, therefore, electrons can easily jump from valence to conduction band, resulting in the resistance to decrease.