
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
218.4k+ 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
A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

ill in the blanks Pure tungsten has A Low resistivity class 12 physics JEE_Main

The value of the resistor RS needed in the DC voltage class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

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

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

