Question

The temperature coefficient of resistance of wire is $12.5\times {{10}^{-4}}/{}^\circ C$ . At 300K the resistance of the wire is 1ohm. The temperature at which resistance will be 2ohm is
A. 1154K
B. 1100K
C. 1400K
D. 1127K

Answer 1

Hint: You could either recall the expression of temperature dependence of resistance or you could simply derive it from the idea that a fractional change in resistivity is proportional to change in temperature. This relation that is in terms of resistivity is made in terms of resistance by assuming length and area doesn’t change with temperature. Now all that is left to do is simple substitution.

Formula used: Expression for temperature dependence of resistance,
$R={{R}_{0}}\left( 1+{{\alpha }_{T}}\Delta T \right)$

Complete step by step answer:
We all know that resistance depends on the material of the conductor as well as the geometry of the conductor. But we often consider the temperature dependence of resistance negligible. Since, the given question is based on the temperature dependence of resistance of the material, let us first discuss the temperature dependency of the material before going into the question.
Remember that the collision process highly influences the electrical resistance of a conductor. Basically, it is the resistivity that shows a linear relationship with temperature. That is,
$\rho \propto \dfrac{1}{d}\propto {{A}^{2}}\propto T$
Where d= mean free path or simply, the average distance travelled between two successive collisions.
A= amplitude of atomic vibrations
T= temperature
Ρ=resistivity
So this proportionality relation can be explained as, when the temperature is increased, average kinetic energy of electrons increases, number of collisions taking place increases, mean free path decreases, it becomes difficult for the electrons to get by. Therefore, the resistivity increases with temperature. Or simply we could say that, a fractional change in resistivity is proportional to change in temperature, that is,
$\dfrac{\Delta \rho }{{{\rho }_{0}}}\propto \Delta T$
$\dfrac{\Delta \rho }{{{\rho }_{0}}}={{\alpha }_{T}}\Delta T$ …………… (1)
Where, ‘${{\alpha }_{T}}$’ is the temperature coefficient of resistivity.
Equation (1) can be rearranged as,
$\rho ={{\rho }_{0}}\left( 1+\alpha \Delta T \right)$ ……………. (2)
We know that, normally, the linear coefficient is much less than the temperature coefficient of resistivity, so we could assume that the length along with the area doesn’t change with the temperature. Thus, we can write (2) in terms of resistance as,
$R={{R}_{0}}\left( 1+{{\alpha }_{T}}\Delta T \right)$ ………………. (3)
Now we could simply substitute the values given in the question in equation (3) and find the answer.
$R=2\Omega $
${{R}_{0}}=1\Omega $
${{\alpha }_{T}}=12.5\times {{10}^{-4}}/{}^\circ C$
${{T}_{0}}=300K$
$2=1\left( 1+\left( 12.5\times {{10}^{-4}}\times \left( T-300 \right) \right) \right)$
$12.5\times {{10}^{-4}}\left( T-300 \right)=1$
$T-300=\dfrac{1}{12.5\times {{10}^{-4}}}=800$
$T=800+300=1100K$

So, the correct answer is “Option B”.

Note: Since the temperature coefficient of the wire is given in /℃ you may think of converting the given temperature into ℃, which is quite unnecessary. We are here dealing with the temperature change, so, the change in temperature remains the same irrespective of which units they are in. Also, remember all materials do not have the same dependence of temperature. (For example, in semiconductors, α is negative.)