Question

The resistance of a wire at $20{}^\circ C$ is $20ohms$ and at $500{}^\circ C$ is $60ohms$. At which temperature its resistance will be $25ohms$ ?
A) $80{}^\circ C$
B) $70{}^\circ C$
C) $60{}^\circ C$
D) $50{}^\circ C$

Answer 1

Hint: Resistance is caused by obstruction in motion of electrons. Temperature is a measure of vibrations of particles of matter. Resistance of a conductor or a wire increases with rising temperature with respect to the constant $\alpha $ which is a constant called as ‘temperature coefficient’.

Formula Used:
The relation of resistance with temperature is as follows,
$RT=RT_1\left[ 1+\alpha \Delta T \right]$
where, $RT$ is the resistance at a given temperature, $RT_1$ is the initial resistance at a given temperature which is the initial temperature, $\alpha $ is the constant which is called the ‘temperature coefficient of resistance of the material of the wire, and $\Delta T$ Is the change in the temperature with respect to the initial temperature and represented as, $\Delta T=T-T_1$ where, $T$ is the temperature for resistance $RT$, and $T_1$ is the temperature for resistance $RT_1$.

Complete step by step answer:
Given: initial resistance $RT_1=20\Omega $ at temperature $T_1=20{}^\circ C$
Resistance at a given temperature $T=500{}^\circ C$ is$RT=60\Omega $.
Now using the formula $RT=RT_1\left[ 1+\alpha \Delta T \right]$ we obtain the value of $\alpha $as follows,
Substituting the given values in the above formula we get,
$60=20\left[ 1+\alpha \left( 500-20 \right) \right]$
$\Rightarrow \dfrac{60}{20}=1+480\alpha \Rightarrow 3-1=480\alpha $
$\Rightarrow \alpha =\dfrac{2}{480}\Rightarrow \alpha =0.00417$
Now let the temperature for a given resistance of $RT_2=25\Omega $ be $T_2$
Also, the initial resistance is $RT_1=20\Omega $ at temperature $T_1=20{}^\circ C$
For resistance $RT_2=25\Omega $ the given formula is obtained from the above-mentioned formula,
$RT_2=RT_1\left[ 1+\alpha \left( T_2-T_1 \right) \right]$ ( Here $T_2-T_1=\Delta T$ which is change in temperature with respect to initial temperature)
Also the value for temperature coefficient of resistance is given by $\alpha =0.00417$
Substituting the given values, we get,
$25=20\left[ 1+0.00417\left( T_2-20 \right) \right]$ On solving we get,
$\begin{align}
  & \Rightarrow \dfrac{25}{20}=1+0.00417\left( T_2-20 \right)\Rightarrow 1.25-1=0.00417\left( T_2-20 \right) \\
 & \Rightarrow \dfrac{0.25}{0.00417}=T_2-20 \\
\end{align}$
$\Rightarrow T_2=59.95+20\Rightarrow T_2=79.95\approx 80{}^\circ C$
Hence the correct option is option (A).

Note: The resistance of the wire on raising the temperature depends on the ‘temperature coefficient of resistance’ of the material of the wire. Also solve the question in the correct order starting from taking the initial resistance as $RT_1=20\Omega $ . Always calculate the values up to two decimal places for getting a correct answer. Calculations should be done accurately.