
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
124.2k+ views
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.
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.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main Login 2045: Step-by-Step Instructions and Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Physics Average Value and RMS Value JEE Main 2025

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
