Question

The resistance of a wire at room temperature ${30^ \circ }C$ is found to be \[10\Omega \]. Now to increase the resistance by $10\Omega $, the temperature of the wire must be: (The temperature coefficient of resistance of the material of the wire is 0.002)
(A) ${36^ \circ }C$
(B) ${86^ \circ }C$
(C) ${560^ \circ }C$
(D) ${33^ \circ }C$

Answer 1

Hint:Here we are going to apply the concept of resistance as a temperature dependent parameter.Resistance increases with increase in temperature (except for semiconductors) as electrons face more collisions on their paths. The resistance linearly depends upon temperature.

Formula used:
\[{R_t} = {R_o}(1 + \alpha T)\]
Where,
${R_t}$is the resistance at required temperature T
${R_o}$is the resistance at the 0 degree celsius
$\alpha $is the temperature coefficient for the given material

Complete step by step answer:
 Generally, the proportionality constant known as resistivity depends upon the temperature. Now, by the formula,
\[R = \rho \dfrac{l}{A}\]
Length and area are independent of temperature so resistance depends upon temperature as well.
With increase in temperature the electrons attain more and more energy in the form of kinetic energy. The increase in energy results in an increase in velocity of electrons which in turn means that the electrons will cover the mean path in less time. Thus, the number of collisions will increase and therefore in time t the effective number of electrons reaching point B from point A will decrease. This reduces the amount of charge reaching from A to B in time t. this is nothing but the reduced current (charge per sec). For a given potential difference if current decreases then resistance increases.
This is the reason that with increase in temperature resistance increases. The relation is given as:
\[{R_t} = {R_o}(1 + \alpha T)\]
According to the question:
For given data (as we do not know temperature at 0 degree celsius), we have,
\[10 = {R_o}(1 + 30\alpha )\]
For present situation,
\[20 = {R_o}(1 + \alpha T)\]
Dividing both the equations, we get,
\[
\dfrac{{20}}{{10}} = \dfrac{{{{{R_o}}}(1 + \alpha T)}}{{{{{R_o}}}(1 + 30\alpha )}} \\
\Rightarrow 1 + \alpha T = 2 + 60\alpha \\
Now, \alpha=0.002 \\
\Rightarrow T = \dfrac{1}{{0.002}} + 60 \\
\Rightarrow T = {560^ \circ }C \\
 \]
Therefore, the correct answer is option C.

Note: It may be the case with insulators and metals that the resistance increases with temperature due to the explained reason. But in semiconductors the given explanation doesn’t work. Resistance decreases with increase in temperature. This is because
Semiconductors with increase in temperature have more holes and electrons are created due to extra energy. Since the number of charge carriers increases the current increases and resistance decreases for given potential difference.Since the number of charge carriers increase the current increases and resistance decreases for given potential difference.