
The resistance of wire in a heater at room temperature is \[65\Omega \] . When the heater is connected to a 220V supply the current settles after a few seconds to 2.8A. What is the steady temperature of the wire. (Temperature coefficient of resistance \[\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}}\] )
(A) \[955^\circ C\]
(B) \[1055^\circ C\]
(C) \[1155^\circ C\]
(D) \[1258^\circ C\]
Answer
232.8k+ views
Hint: We will first find the resistance when the heater is connected to a 220V supply using the formula \[R = \dfrac{{Supply\,voltage}}{{Supply\,current}}\] .
Then we use the formula of dependence of resistance with temperature i.e. \[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] as the resistance vary with the temperature.
After that we will calculate the temperature by putting all the values given in the question.
Complete step by step solution
It is given resistance of wire in a heater at room temperature i.e. \[{R_1} = 65\Omega \] and temperature is given i.e. room temperature \[{T_1} = 27\;^\circ C\] .
Now when the heater is connected to a 220V supply resistance \[{R_2} = \dfrac{{Supply\,voltage}}{{Supply\,current}}\] , now we put the values as given in question.
So \[{R_2} = \dfrac{{220V}}{{2.8A}} = 78.6\Omega \]
Now, using the relation
\[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] , where \[{R_2}\] is the resistance at temperature \[{T_2}\] , \[{R_1}\] is the resistance at temperature \[{T_1}\] and \[\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}}\] .
we can find \[{T_2} - {T_1}\] from this equation as we know the value of \[{R_2},{R_1}\] and \[\alpha \] .
So, we solve this equation:
\[{R_2} = {R_1} + {R_1}\alpha ({T_2} - {T_1})\]
\[{T_2} - {T_1} = \dfrac{{{R_2} - {R_1}}}{{{R_1}}} \times \dfrac{1}{\alpha }\] , now we put the values and find the value of \[{T_2} - {T_1}\]
\[{T_2} - {T_1} = \dfrac{{78.6 - 65}}{{65}} \times \dfrac{1}{{1.7 \times {{10}^{ - 4}}}}\] =1231
So \[{T_2} = 1231 + {T_1}\] and we know that \[{T_1} = 27\;^\circ C\]
So, after putting value of \[{T_1}\] in \[{T_2}\] we get:
\[{T_2} = 1258\;^\circ C\] .
So, option D is correct.
Note: Always remember that we will take the temperature of room \[27\;^\circ C\] and also the formula of resistance when a conductor is connected with supply voltage.
Also remember the general rule i.e. resistivity increases with increasing temperature in conductors and decreases with increasing temperature in insulators.
Then we use the formula of dependence of resistance with temperature i.e. \[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] as the resistance vary with the temperature.
After that we will calculate the temperature by putting all the values given in the question.
Complete step by step solution
It is given resistance of wire in a heater at room temperature i.e. \[{R_1} = 65\Omega \] and temperature is given i.e. room temperature \[{T_1} = 27\;^\circ C\] .
Now when the heater is connected to a 220V supply resistance \[{R_2} = \dfrac{{Supply\,voltage}}{{Supply\,current}}\] , now we put the values as given in question.
So \[{R_2} = \dfrac{{220V}}{{2.8A}} = 78.6\Omega \]
Now, using the relation
\[{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right]\] , where \[{R_2}\] is the resistance at temperature \[{T_2}\] , \[{R_1}\] is the resistance at temperature \[{T_1}\] and \[\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}}\] .
we can find \[{T_2} - {T_1}\] from this equation as we know the value of \[{R_2},{R_1}\] and \[\alpha \] .
So, we solve this equation:
\[{R_2} = {R_1} + {R_1}\alpha ({T_2} - {T_1})\]
\[{T_2} - {T_1} = \dfrac{{{R_2} - {R_1}}}{{{R_1}}} \times \dfrac{1}{\alpha }\] , now we put the values and find the value of \[{T_2} - {T_1}\]
\[{T_2} - {T_1} = \dfrac{{78.6 - 65}}{{65}} \times \dfrac{1}{{1.7 \times {{10}^{ - 4}}}}\] =1231
So \[{T_2} = 1231 + {T_1}\] and we know that \[{T_1} = 27\;^\circ C\]
So, after putting value of \[{T_1}\] in \[{T_2}\] we get:
\[{T_2} = 1258\;^\circ C\] .
So, option D is correct.
Note: Always remember that we will take the temperature of room \[27\;^\circ C\] and also the formula of resistance when a conductor is connected with supply voltage.
Also remember the general rule i.e. resistivity increases with increasing temperature in conductors and decreases with increasing temperature in insulators.
Recently Updated Pages
JEE Main 2026 Helpline Numbers - Center Contact, Phone Number, Address

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

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

