The temperature coefficient resistivity of a material is $0.0004/K$. When the temperature of the material is increased by ${50^\circ }C$ , its resistivity increases by $2 \times {10^{ - 8}}$. The initial resistivity of the material of the resistance is:
A) $50 \times {10^{^{ - 8}}}$
B) $90 \times {10^{ - 8}}$
C) $100 \times {10^{ - 8}}$
D) $200 \times {10^{ - 8}}$
Answer
249.3k+ views
Hint: In the given question, we need to find the initial resistivity of the material given the rise in resistivity due to the increase in temperature, this can be done using the relation, as we know, and the initial resistivity is directly proportional to the rise in temperature.
Formula used:
${\rho _t} = {\rho _0}[1 + \alpha \Delta T]$
where,
${\rho _t} = $ Resistivity at standard temperature
${\rho _0} = $Initial resistivity
$\alpha = $Coefficient of resistivity
$\Delta T = $Rise in temperature
Complete step by step solution:
Resistivity can be defined as the resistance of a conductor having unit length and unit cross-sectional area. This is a characteristic property of the material, and thus it is directly proportional to temperature.
As temperature increases, there is a rise in resistivity, similarly, with a decrease in temperature, the resistivity of the conductor also decreases.
Thus, we can say that the resistivity $f$ a conductor depends on the material and temperature but is independent of the shape and size of the conductor.
We can say, $\rho \propto \Delta T$
where,
$\rho = $Resistivity of the material
$\Delta T = $Rise in temperature
We know the formula that states the relation between resistivity and rise in temperature goes as:
${\rho _t} = {\rho _0}[1 + \alpha \Delta T]$
where,
${\rho _t} = $ Resistivity at standard temperature
${\rho _0}= $Initial resistivity
$\alpha = $Coefficient of resistivity
$\Delta T = $Rise in temperature
On rearranging the equation, we obtain:
$\Rightarrow {\rho _t} - {\rho _0} = {\rho _0}\alpha \Delta T$
We know:
$\Rightarrow {\rho _t} - {\rho _0}= $Rise in resistivity
In the question, we have the value for the change in resistivity,
Putting the values, in the equation:
$\Rightarrow 2 \times {10^{ - 8}} = {\rho _0} \times 0.0004 \times 50$
$ \Rightarrow {\rho _0} = \dfrac{{2 \times {{10}^{ - 8}}}}{{0.0004 \times 50}}$
Thus, on solving the equation we obtain:
$\Rightarrow {\rho _0} = 100 \times {10^{ - 8}}ohm - m$
This is the required solution.
Thus, option (C ) is correct.
Note: Although, with an increase in temperature resistivity increases however resistance decreases. This happens because, as the temperature rises, the band gap between conduction and valence band decreases, therefore, electrons can easily jump from valence to conduction band, resulting in the resistance to decrease.
Formula used:
${\rho _t} = {\rho _0}[1 + \alpha \Delta T]$
where,
${\rho _t} = $ Resistivity at standard temperature
${\rho _0} = $Initial resistivity
$\alpha = $Coefficient of resistivity
$\Delta T = $Rise in temperature
Complete step by step solution:
Resistivity can be defined as the resistance of a conductor having unit length and unit cross-sectional area. This is a characteristic property of the material, and thus it is directly proportional to temperature.
As temperature increases, there is a rise in resistivity, similarly, with a decrease in temperature, the resistivity of the conductor also decreases.
Thus, we can say that the resistivity $f$ a conductor depends on the material and temperature but is independent of the shape and size of the conductor.
We can say, $\rho \propto \Delta T$
where,
$\rho = $Resistivity of the material
$\Delta T = $Rise in temperature
We know the formula that states the relation between resistivity and rise in temperature goes as:
${\rho _t} = {\rho _0}[1 + \alpha \Delta T]$
where,
${\rho _t} = $ Resistivity at standard temperature
${\rho _0}= $Initial resistivity
$\alpha = $Coefficient of resistivity
$\Delta T = $Rise in temperature
On rearranging the equation, we obtain:
$\Rightarrow {\rho _t} - {\rho _0} = {\rho _0}\alpha \Delta T$
We know:
$\Rightarrow {\rho _t} - {\rho _0}= $Rise in resistivity
In the question, we have the value for the change in resistivity,
Putting the values, in the equation:
$\Rightarrow 2 \times {10^{ - 8}} = {\rho _0} \times 0.0004 \times 50$
$ \Rightarrow {\rho _0} = \dfrac{{2 \times {{10}^{ - 8}}}}{{0.0004 \times 50}}$
Thus, on solving the equation we obtain:
$\Rightarrow {\rho _0} = 100 \times {10^{ - 8}}ohm - m$
This is the required solution.
Thus, option (C ) is correct.
Note: Although, with an increase in temperature resistivity increases however resistance decreases. This happens because, as the temperature rises, the band gap between conduction and valence band decreases, therefore, electrons can easily jump from valence to conduction band, resulting in the resistance to decrease.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Centrifugal Force in Physics

