Question

At what temperature will the resistance of a copper wire become three times its value of ${{0}^{\text{o}}}C$ ? (Temperature coefficient of resistance for copper = $4\times {{10}^{-3}}pe{{r}^{\text{o}}}C$):
$\left( a \right){{500}^{\text{o}}}C$
$\left( b \right){{450}^{\text{o}}}C$
$\left( c \right){{600}^{\text{o}}}C$
$\left( d \right)\text{None of the above}$

Answer 1

Hint: As the question directly informs us about the temperature coefficient of resistivity of a metal (here copper) wire, we will use the direct formula of it and substitute required values in it. We will focus on solving the difference between temperature of a conducting material and its specific value.

Formula used:
${{R}_{T}}={{R}_{\text{o}}}\left( 1+\alpha \left( \Delta T \right) \right)$ where $\Delta T$ is the difference between temperature of conductor and the temperature at specific value of same substance, ${{R}_{T}}$ and ${{R}_{\text{o}}}$ are variations in resistance and $\alpha $ is temperature coefficient of resistance for conducting material.

Complete answer:
Temperature coefficient resistance: If we are getting some changes in values of resistance with the change of temperature in degrees then we call that observed value as a temperature coefficient resistance. We calculate for temperature coefficient by using the formula ${{R}_{T}}={{R}_{\text{o}}}\left( 1+\alpha \left( \Delta T \right) \right)$.
As the value of resistance given to us, of a copper wire is three times its value so this means that ${{R}_{T}}=3{{R}_{\text{o}}}$,
Value of $\alpha $ for conducting material is $\alpha =4\times {{10}^{-3}}$.
Now, we will use the formula ${{R}_{T}}={{R}_{\text{o}}}\left( 1+\alpha \left( \Delta T \right) \right)$ and substitute the values, so we get
$3{{R}_{\text{o}}}={{R}_{\text{o}}}\left( 1+\left( 4\times {{10}^{-3}} \right)\left( \Delta T \right) \right)$
$\Rightarrow 3=1+\left( 4\times {{10}^{-3}} \right)\left( \Delta T \right)$
$\Rightarrow 3-1=4\times {{10}^{-3}}\left( \Delta T \right)$
$\Rightarrow \Delta T=\dfrac{2}{4\times {{10}^{-3}}}$
$\Rightarrow \Delta T=\dfrac{1}{2\times {{10}^{-3}}}$
$\Rightarrow \Delta T=\dfrac{1000}{2}$
$\Rightarrow \Delta T=500$
Hence, the correct option is $\left( a \right){{500}^{\text{o}}}C$.

Note: There are two types of it namely, positive temperature coefficient and negative temperature. By the positive temperature resistance, we mean the increment in the resistivity and resistance of any substance which is proportional to the decrease in relaxation time of collisions. On the other hand, the negative temperature coefficient, there is an increase in the number of carriers of charge per unit of volume which increase with the increase in temperature. There are many materials whose value related to this coefficient is fixed. But if value of change in temperature is given to us along with the value of change in resistances then we will apply the formula ${{R}_{T}}={{R}_{\text{o}}}\left( 1+\alpha \left( \Delta T \right) \right)$ and solve further.